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I'm not sure this is generically true, but as part of something I'd like to prove I'm hoping to make use of something along the lines of being able to write for some symmetric, nonzero $B$

$$ B = A B A^\intercal $$

and using that to infer that $A$ must be the identity matrix

I think I can infer that $A$ is a similarity transform, and therefore unitary, and I know that I can write

$$ B = (A^n) B (A^\intercal)^n $$

for any positive integer $n$, but beyond that all I know is that I'm trying to find the centralizers of this matrix in the group of unitary matrices

b3m2a1
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  • If $A$ works then so does $-A$ so the solution cannot be unique. – copper.hat Sep 29 '23 at 19:24
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    If $B$ symmetric (with real entries) then there is an orthonormal basis consisting of eigenvectors of $B,$ say ${v_k}_{k=1}^n$ Then $Av_k=a_kv_k$ is unitary for $a_k=\pm 1$ and $B=ABA^T.$ Hence for any symmetric matrix $ B$ there is $A,$ which is not a multiple of $I,$ such that $B=ABA^T.$ – Ryszard Szwarc Sep 29 '23 at 19:31
  • @copper.hat fair, I should probably have asked for conditions where $A = \lambda I$ for some scalar – b3m2a1 Sep 29 '23 at 19:33

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