If you see a word in quotes for no reason, it's me making a definition of that term that I will use later.

Counting circles all lie on 6 lines (4 forming a "hash" in boxes 1379, and 2 forming a "cross" though box 5), so values over 6 are impossible.
Summing to 20 means the values are 23456.
6 must appear in both row 5 and column 5

(You can colour cells that can't contain 6 if you want, it's a lot of cells, and rejecting 6 is useful quite often. You could instead colour the hash, as those cells are constantly being referred to. But this isn't "colouring" in the usual sense, just a visual aid.)

The sum of the 8 cells in box 5 is twice the sum of the circles connecting them.
=> r5c5 is odd, but not 5 because 5 must appear in at least row 5 or column 5
r5c5 cannot be 1, as 44=22*2 requires two 6s and two 5s in the cirles, and you simply can't fit a 6 into box 5 if there are two 6s in the circles outside.
r5c5 cannot be 3 as 42=21*2 requires at least two 6s or three 5s, which again is impossible.
Note that the reason you can't place two 5s diametrically opposite each other in the inner circles, as there's nowhere for a 5 in box 5, applies to any other number too, the combination of DAs and sudoku reject that number from every cell in box 5 apart from the very centre (and 7 and 9 can't go on circles).

r5c5 being 7 would leave 38=19*2, and 9 would leave 36=18*2.

The 4 hash lines contain 16 circles, 8 "inner" ones in boxes 2468, and 8 "outer" ones in boxes 1379.

It's instructive to know how many of the four required 6s are in the inner circles, and how many in the outer circles.

At this point, a wise person would say "the symmetry breaking of the furniture in boxes 1379 looks like it will be an annoyance - let's solve the simpler symmetric puzzle first, and worry about rotating it into place later". Do it. You know it makes sense. It takes 10 seconds to draw the 5 lines in and 16 circles in sudokupad. Do it.

If all four 6s were in the outer circles, the central star would need to contain 5-5-4-4-.
The DAs wholly in boxes 2468 ("spurs") all have an inner end, and a cell midway along their line, all of which see a 4 and a 5, either in their own box, or in the box opposite.
Making the DAs work restricts their outer circles, and you're left with two pairs of ([23])-[789]-([46]) and ([23])-[789]-([56]), where 6 is represented in both, and 5 in at least one.
The 5 in box 5 is now obliged to be on the 4--4 DA with a 3, and the 4 is on the 5--5 DA with a 6, opposite the 5 - blocking the [46] and [56] endpoints of a spur DA pair.
Therefore at least one of the 6s must be in the inner circles.

So we need to try and fit 6s into the inner circles.

If you put a 6 on one of the circles in the central star, it forces a 6 into the opposite corner of box 5 and onto the cross in the opposite box, which can now only be on a (6)-[89]-([23]) DA.
Its neighbours round the star can be 552, 553, 544, or 543.
If its neighbours are 55x, then one of the 5s must be opposite the 6.
552 can be rejected both ways round:
If the DA leading straight out of the 6 joins to the 2, the 5--5 DA is (5)-6-4-(5), and the 5--2 DA is unsatisfiable.
If the DA leading straight out of the 6 joins to the other 5, the DA opposite is (5)-6-1-(2) that forces a (2)-3-5-(6) DA, knocking a 5 out of one of the cross lines, and the 235 all in a line force the DA it points towards to have [46] at one end, and thus forcing a 5 onto the other cross line. However, we've already put a 5 in that box.
553 likewise:
If the DA leading straight out of the 6 joins to the 3, the 5--5 DA is again (5)-6-4-(5), the 5--3 is (5)-[17]-[17]-(3), and the 3--6 DA is unsatisfiable.
If the DA leading straight out of the 6 joins to the 5, the 5--3 DA is (5)-6-2-(3), the 3--6 DA is (3)-4-5-(6), the 6--5 DA is (6)-8-3-(5), and the spur DA perpendicular is unsatisfiable.
544 requires a 7 in r5c5, but that doesn't leave enough high digits to satisfy the DAs in box 5.
543 has more cases to consider:
Placing the 3 opposite the 6 makes the two DAs connecting to each other clash.
Placing the 3 at the end of the DA leading straight out of the 6 leaves 2 cases:
If the central star goes 6-3-4-5, then the 3--4 DA is (3)-[25]-[25]-(5), and that 2 forces the spur DA in the box with the 5 to begin with a 5, and then be unsatisfiable.
If the central star goes 6-3-5-4, then the 4--6 DA is (4)-[28]-[28]-(6), and the 6--3 DA is (6)-5-4-(3), and the 5 and the [28] pair make the spur DA in the box with the 3 unsatisfiable.
Therefore, if the central star completes 543, the 3 is clockwise of the 6.
If it's 6543, then the 4--3 DA is (4)-6-1-(3), the 3--6 DA is (3)-4-5-(6), the 6--5 DA is (6)-8-3-(5), and the 5--4 DA is (5)-[27]-[27]-(4). A 6 can no longer be placed on either the spur DA perpendicular to the 6--5 on the central star, nor the one opposite. Fail.
If it's 6453, then the whole central box and all spur DAs can be resolved uniquely - however, to do so requires use of three 2's in counting circles.
Therefore, there's no 6 in the central star's circles.
As it must sum to 18, and can't have a tripled value, it must be 5544, and r5c5 is 9.

Get yourself a cup of tea - you've deserved it!

So, we now need to place a 6 we know exists in the inner circles relative to the 5544 in the inner circles. If you've not taken my advice about solving the simple symmetric puzzle first, correct your error. Throw away the furniture, break symmetry, and hypothesise a placement of the 5s and the 4s on the central start. Do it. Now.

The 4 must go with a 6 on the DA between the 5s, and the 5 must go with a 3 on the DA between the 4s.
A 6 must go at the far end of the spur behind the 5--5 DA.
A 6 must go at the near end of the spur behind the 4--4 DA - that's the one we were looking for. Too easy!
That however means that we've found the line on the cross with just a 6, so the other line on the cross must have both a 6 and a 5 at its ends. There's only one way to do that - the two 4s point towards the 5, and the two 5s point towards the 6, whose DA becomes (2)-8-(6).
It's worth noting that there are only 4 circles available to accept the final two 4s, one of which is forbidden in any combo, and one of which is obligated, which also puts a 4 in the crook of the 286 we've just found.
If the 4s are placed on the symmetric diagonal rather than as a rotational pair, you simply can't place the 6s any more, so they're rotational, 2 steps away from the 5s on the inner star.
That forces the three remaining circles 6s to be rotationally placed next to those two 4s (touching the 2 inner 5s), and in the only remaining possible circle.
There are now only two non-clashing places for the two remaining circled 5s.

There's not much we can conclusively do in the simpler puzzle apart from filling the spur DA behind the 5--5 DA with (3)-9-(6), because the [18] or [27] pair in box 5 must have either a 2 or an 8, so (2)-8-(6) is impossible. There's also a naked 4 next to that 6, and a naked 5 rotationally opposite it

However, some pencil markings are still useful, as we can try to see which bits of furniture would fit where (all 4 are rotationally distinct).

Of particular interest is the corner with [23] and 5 in the circles, as either of the 5-cell DAs would be quite tricky to fit in. The [23] couldn't be a 2, because 1+3+4>7; and couldn't be a 3 because 1+2+4<8 and 1+2+6>8.

Therefore the [23]/5 corner is top left or bottom right in the orignal, and has one of the 4-cell simple angular DAs on the circles.

The crooked 5-cell DAs are quite hard to satisfy too, but still leave choices.
The one terminating 64 must contain [235] or [127] - definitely a 2.
The one terminating 65 must contain [137] or [128] - definitely a 1.

However, the box with the 6---5 DA must have a 4 in one of only two cells, and one of them is on both orientations of the line, a line that cannot contain a 4.
So we've found the 4 in that cell, and worked out which of the lines is in which corner.

Go back to the original puzzle, and copy over what you've learnt.

Now do a bit of sudoku!
box 9 gets a 4 and a 6 instantly, and box 8 gets its 6 too.
however, box 9's DA gets a [12], so it's either 2 with 1, or 3 with 2. So wipe the 2s from row 8.
Box 1's DA can't be (4)-5-5-(6), so box 1's 5 is in the corner.
=> box 7's DA has a 5 next to the 6, and a [23] pair.

We still have three counting circles unresolved, one needs to be a 2, the other two 3s.
If box 9 were to have the 2, box 8's 3, and box 7's [23] pair mean that there's nowhere for box 9's 3. So box 9 has a 3 on the circle.
We can't place the 2 or 3 in box 8 yet, but it's a [23] pair in row 7.

We know the wonku DA in box 3 has filling either [128] or [137].
If it was [128], then 3 goes in column 9, column has a [79] pair, box 9 turns into a [18] pair, and box 6 is stranded with a 2 in column 7 ... which puts another 3 in column 9.
=> box 3's DA is [137], which leaves an [89] pair in row 3
=> r1c7 is 2
=> r2c2 is 2 and r1c2 is 8
=> column 3's 2 is in r5c3, also placing the 8 in box 4, and resolving that row.
=> r5c9 is 3, r6c8 in the DA is 8
=> r2c8 is 3, leaving a [17] pair in column 9, making an [89] pair, and placing the 2 in r6c9
=> the [78] pair in box 6 and the [18] pair in box 5 both resolve the [17] pair in box 4.
=> The DA in box 1 must be 8 above 2, and the 3 goes in r3c1
=> box 2's 3 is in r1c4
=> box 7's [23] pair resolves 2, 3.
=> box 8's [23] pair resolves 2, 3.
=> there's a 7 on the DA at r8c4
=> r8c5 is 8, r6c5 is 1, r6c4 is 8, r3c5 is 7, r3c3 is 1

Placing a 9 in r8c1 pushes it round r1c3 and r2c4 into r8c6
=> r8c1 is 1, and let's speedrun!
r8c6 is 9, r1c6 is 1, r2c4 is 9, r2c1 is 7, r1c3 is 9, r2c9 is 1, r1c9 is 7, r7c1 is 9, r7c9 is 8, r9c8 is 7, r9c3 is 8, r7c3 is 7, r9c4 is 1, r3c9 is 9, r3c7 is 8, r5c7 is 7, r5c8 is 8, r7c7 is 1, r9c7 is 9

And Robert's your auntie's live in lover!

Proof of uniqueness:
https://sigh.github.io/Interactive-Sudoku-Solver/?q=.CountingCircles%7ER1C5%7ER2C3%7ER3C2%7ER5C1%7ER7C2%7ER8C3%7ER9C5%7ER8C7%7ER7C8%7ER5C9%7ER3C8%7ER2C7%7ER3C4%7ER4C3%7ER6C3%7ER7C4%7ER7C6%7ER6C7%7ER4C7%7ER3C6.DoubleArrow%7ER4C7%7ER4C6%7ER4C5%7ER3C4.DoubleArrow%7ER3C4%7ER4C4%7ER5C4%7ER6C3.DoubleArrow%7ER6C3%7ER6C4%7ER6C5%7ER7C6.DoubleArrow%7ER7C6%7ER6C6%7ER5C6%7ER4C7.DoubleArrow%7ER3C6%7ER2C6%7ER1C5.DoubleArrow%7ER4C3%7ER4C2%7ER5C1.DoubleArrow%7ER7C4%7ER8C4%7ER9C5.DoubleArrow%7ER6C7%7ER6C8%7ER5C9.DoubleArrow%7ER2C7%7ER2C8%7ER1C9%7ER2C9%7ER3C8.DoubleArrow%7ER7C2%7ER8C2%7ER9C1%7ER9C2%7ER8C3.DoubleArrow%7ER3C2%7ER2C2%7ER1C2%7ER2C3.DoubleArrow%7ER8C7%7ER8C8%7ER8C9%7ER7C8

Midway point (with arbitrarily-chosen wrong rotation):
https://sigh.github.io/Interactive-Sudoku-Solver/?q=.CountingCircles%7ER1C5%7ER2C3%7ER3C2%7ER5C1%7ER7C2%7ER8C3%7ER9C5%7ER8C7%7ER7C8%7ER5C9%7ER3C8%7ER2C7%7ER3C4%7ER4C3%7ER6C3%7ER7C4%7ER7C6%7ER6C7%7ER4C7%7ER3C6.DoubleArrow%7ER4C7%7ER4C6%7ER4C5%7ER3C4.DoubleArrow%7ER3C4%7ER4C4%7ER5C4%7ER6C3.DoubleArrow%7ER6C3%7ER6C4%7ER6C5%7ER7C6.DoubleArrow%7ER7C6%7ER6C6%7ER5C6%7ER4C7.DoubleArrow%7ER3C6%7ER2C6%7ER1C5.DoubleArrow%7ER4C3%7ER4C2%7ER5C1.DoubleArrow%7ER7C4%7ER8C4%7ER9C5.DoubleArrow%7ER6C7%7ER6C8%7ER5C9.%7ER3C4_5%7ER4C7_5