Overview:
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- find the weak point
- slow and steady flow
- speedrun

Find the Weak Point
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Clearly, renbans and whispers cannot solve for polarity, therefore the 3 in the corner does that job and we have to solve in A-I until we can assert C is 3 or G is 3.
There's not enough pressure to correlate polarities along different whispers.
Renbans don't intrinsically have to preserve polarity, as they could be a 5 between D and F.
However, the D must be flanked with two Is, and the F with two As.
Colouring the 2nd and 3rd whispers along, you can see that one of the renbans must have such a flip.
And the 5 can't be on r3c5.
=> r7c3 is 5, and arbitrarily label r6c3 D, and r7c4 F, flanked with Is and As.

And you're off!

Flow
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Unify the colouring on those two lines, and label with [ABCD] or [FGHI] as appropriate
r2c4, r5c5, r4c6, and r3c6 are more restricted, not being able to be D or F.
The renban out of r6c5's A must be a [BC] pair in row 7.
Likewise the renban out of r5c3 must be a [FGH] triple.
Colour and label the first and fourth whisper, now one cell's polarity is known.
=> r5c2 is 5, and complete the colouring on column 2.
The renban thought r3c5 must have an A or D there, but isn't A, so is D.
A [GH] pair in column 5 restricts r5c1 to [FI], but it neighbours [BC] on a whisper, so must be I
A [GH] pair in row 3 makes r3c2 F.
A [BC] pair in column 6 makes r9c6 D, and along the whisper r8c6 is I
=> the [5I] pair in r1c1/r2c1 resolve 5 above I
The [BC] pairs r2c4, r4c6, r7c6, r7c7 alternate, so r2c7 sees both and is A
=> Box 2's A is in r1c6 
=> column 2's A is in r9c2
The renban thru' r3c7 connects [GH] to [GHI], and can't have an F, so has I as the connector
=> box 3's 5 is in r3c8
Because r2c9 and r2c2 are a [GH] pair because of their whispers, r2c3 is D.
Because r4c2 and r4c8 are a [GH] pair, r4c1 is F
=> r3c1 is A
=> r4c3 is A
column 5 needs a [BC] in box 5 and other in box 8, so the former is at r4c5
it also needs an F which must be in r2c5
r5c4 has become a naked D. oo-errr misses.
=> in column 4, r9c4 has become a [GH]

[GH] and [BC] now abound in boxes 1-3.
One of r2c2 or r2c9 is G
=> one of r1c3 or r3c9 is a B
=> one of r1c4 or r2c8 is a C
=> r2c4 is not a C, it's a B

Which is huge!

=> r1c4 and r2c8 are C
=> r3c3 and r3c9 are B
=> r2c8 is a C
=> r8c2 is a C
=> Along the renban, r4c6 has become C
=> Along the whisper it's flanked by Hs which ripples everywhere
Not least, it places Cs in rows 1 and 9, meaning C can't be 3
However, you've come so far, you might as well ljust solve in A-I and fixup at the end.

A [BCG] triple in column 1 makes r8c1 D, and e7c1 H
row 8's H can only be in r8c8
column 7's D can now only be in r4c7
box 6's 5 must be in column 9, so r9c7 is 5

We've still not worked out which of the corners is the 3 (G), but that's simple now
A B in r9c1 puts an [FI] pair in box 9, which puts the G in r8c9, breaking r1c9
=> The G in the corner is in r9c1

Speedrun!
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Just do it - BOSH!


Proof of uniqueness:
https://sigh.github.io/Interactive-Sudoku-Solver/?q=.Whisper%7E5%7ER4C1%7ER3C1%7ER2C2%7ER1C2.Whisper%7E5%7ER8C1%7ER7C2%7ER6C2%7ER5C3%7ER4C3%7ER3C4%7ER2C4%7ER1C5.Whisper%7E5%7ER9C3%7ER8C4%7ER7C4%7ER6C5%7ER5C5%7ER4C6%7ER3C6%7ER2C7%7ER1C7.Whisper%7E5%7ER2C9%7ER3C9%7ER4C8%7ER5C8%7ER6C7%7ER7C7%7ER8C6%7ER9C6.Renban%7ER6C2%7ER7C3%7ER7C4.Renban%7ER2C4%7ER3C5%7ER4C6.Renban%7ER3C6%7ER3C7%7ER4C8.Renban%7ER6C5%7ER7C6%7ER7C7.Renban%7ER5C3%7ER4C2%7ER3C2%7ER2C2.ContainAtLeast%7E3%7ER9C1%7ER1C1%7ER1C9%7ER9C9