"Glitzen Glitzen, Stiller, Stern"

Appologies to German speakers, but the name, like the puzzle itself, is intended to invoke thoughts the Christmas.

Because of the black kropkes, 5s are placed in all corner boxes instantly, severely restricted in the edge boxes, and the white kropkes force two of those edges to have central 5s, due to lack of legal neighbours - that forces the 5 into one of the corner cells of box 5.

The parity of the 4 stars must alternate, as you can't place 2 monogamous numbers in neighbouring cores, you either clash on row or intermediate box.

Therefore you can colour about half of the grid - don't miss all the deductions in box 8.
Now we can solve using [ABCD]/5/[EFGH] until the letters resolve hi-lo. Arbitrarily, the core of the star in box 1 shall be ABCD, and its halo EFGH.
Pencil the possible locations for the monogamous Ds and Fs in boxes 1/3/7/9.
Note also that the two corona cells outside those boxes can not themselves be monogamous, as the two core cells they connect to see each other. Mark all coronas [ABC] or [GHO].

Note the following interesting dependency loops:
A G in r1c2 whispers [AB] into row 2 and forces a C into r2c9. That's a rotation and a parity flip, so continues round the same way. r9c8 has its [HI] shot down, becomes G, and fires an [AB] blocker to r8c1, which becomes C, and fires an [HI] blocker up to r1c2 where we started
Therefore, if any of r1c2, r2c9, r9c8, or r8c1 are G or C, they all are, else none are.
Similar G/C loops connect r2c4, r9c3, r8c6, and r1c7; likewise r4c3, r3c6, r6c7, and r7c4; and finally r3c1, r6c2, r7c9, and r4c8.

The F in the halo in box 9 can only be r9c9 or r9c7.
If it were in r9c7, then box 3's F would be in r3c8, and A is forced into columns 8 and 9. However, the white kropke on row 9 pushes G into r9c6 which must be identical to r7c9. That final G can no longer connect to the [AB] pair that it needs along the whisper.
=> r9c9 is F
=> r9c8 cannot be H because it would interfect with the kropke

If we hypothecate B or C into r3c6, whispers push the F into r3c8 and there's no room for F in box 6
=> r3c6 is A
=> r6c7 is not G, r7c4 is not C, and r4c3 is not G by loop-following.
=> r3c8 cannot be F, as it pushes 4 of G, H and I into column 7.

If we hypothecate H into either r2c7 or r3c7, F goes in the other, and the [GI] pair left clashes with r9c8.
=> the H in box 3 is in column 8, r2c8 or r3c8.

If we hypothecate G into r7c9, I is forced into r9c8, H into r9c7, and I into r6c7, leaving no I in box 3
=> r7c9 is not G.
=> r4c8 is not C, r3c1 is not G, and r6c2 is not C by loop-following.

If we hypothecate G into r8c6, by the loop logic aboves a C goes into r9c3. However, it also prevents D along its whisper, putting box 9's D into r7c8, box 7's into r9c1, and box 8's into r8c4 where it must pair with a C on the kropke in r9c4, putting 2 Cs on row 9.
=> r8c6 is not G.
=> r1c7 is not C, r2c4 is not G, and r9c3 is not C by loop-following.

If we hypothecate I into r1c3, H goes into r3c1, G into r1c2, and H into r4c1. There's nowhere for the D in box 1 now.
=> r1c3 is not I
=> r9c7 is not I equivalently in box 9
=> r7c1 is not A equivalently in box 7

If we hypothecate H or I into r1c1, then r3c1 becomes the other, and r1c2 becomes G. Loop-following puts C in r8c1, and then 2 Fs are needed in column 3.
=> r1c1 is not H or I.
=> r9c1 is not A or B using the same technique.
=> r1c9 is not A or B using the same technique.

Box 3 now has a [BCD] triple - and this is big!
A is forced into r1c7, and B into r4c6.
Whispers put the F in r2c7
H cannot be in r2c8, as the [GI] pair it pits in two 3 clashes with r3c1
=> H is in r3c8.
=> I is in r3c1, and H is in r2c4.
=> G is in r3c7, and I is in r2c8.

... In box 9,  r9c8 is G, r9c7 is H, r9c6 is I, r8c6 is H, r6c7 is I, and r7c9 is I
=> D is in r7c8, A is in r8c8.

... Turning attention to box 7, D is forced into r9c1
=> r9c3 is A and r6c2 is B.
=> r8c3 is F
=> the only place left for I is r8c2.

... In box 1, r1c1 is the only place for F, r2c2 is the only place for A, and r3c2 the only place for D
=> the D in box 3 is in r1c9.

... Down to box 8, the kropke can only be D above C, and r7c4 can only be A, also placing the B in r9c5.
r8c5 is also a naked G, leaving just a [5F] pair.

... Across to box 4, r4c3 is I
Column 9 contains an [AGH] triple! Finally we can look at the blue twinkles usefully
[BC] is either [23] or [78], so can partner with either [146] or [4], which would be [ADF] or [F]
=> B is 2, so change all the letters to numbers - I told you this was huge!

The rest is basically a nobrain speedrun.

Proof of uniqueness:
https://sigh.github.io/Interactive-Sudoku-Solver/?q=.Whisper%7E5%7ER1C2%7ER2C2%7ER3C1%7ER3C2%7ER4C3.Whisper%7E5%7ER1C2%7ER2C3%7ER2C4%7ER3C3%7ER4C3.Whisper%7E5%7ER1C7%7ER2C7%7ER3C6%7ER3C7%7ER4C8.Whisper%7E5%7ER1C7%7ER2C8%7ER2C9%7ER3C8%7ER4C8.Whisper%7E5%7ER6C2%7ER7C2%7ER8C1%7ER8C2%7ER9C3.Whisper%7E5%7ER6C2%7ER7C3%7ER7C4%7ER8C3%7ER9C3.Whisper%7E5%7ER6C7%7ER7C7%7ER8C6%7ER8C7%7ER9C8.Whisper%7E5%7ER6C7%7ER7C8%7ER7C9%7ER8C8%7ER9C8.WhiteDot%7ER3C4%7ER4C4.WhiteDot%7ER4C6%7ER4C7.WhiteDot%7ER7C6%7ER6C6.WhiteDot%7ER6C3%7ER6C4.WhiteDot%7ER9C7%7ER9C6.WhiteDot%7ER8C4%7ER9C4.BlackDot%7ER6C1%7ER7C1