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My question seems like a basic linear algebra problem but I wasn't able to find a solution online nor to come up with one in a short period of time, so here it goes:

Q: What are the equivalence classes of $\mathrm{SU}(n)$ under the equivalence relation $A \sim M A M^T$ for some $M \in \mathrm{SU}(n)$?

This feels like an easy question, yet I am happy to receive some sub-complete answers: What are the general procedures to find all equivalence classes? Are there finitely many, or a "nice" parametrization (similar to eigenvalues/maximal tori for the usual conjugation), etc.

Thanks in advance.

EDIT: Answer found.

The answer is remarkably simple, and is an easy consequence of the final theorem (Theorem 7) in the more general work by Alpin and Ikmarov "A criterion for unitary congruence between complex matrices" from 2012.

Specifically, $A,B \in \mathrm{SU}(n)$ are unitary congruent (i.e. there is an $M \in \mathrm{SU}(n)$ such that $A = M B M^T$) if and only if $\mathrm{Tr}( (A \overline{A})^k) = \mathrm{Tr}( (B \overline{B})^k)$ for all $k=1,\dots, n^2 -1$ (apparently a lower bound there also works but this one seems more intuitive).

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  • $\begingroup$ As it is stated it seems that you are considering matrices up to base change given by the unitary matrix formed of the rows or columns of $M$. Thus, $A\sim B$ if and only if they represent the same linear map. Or am I missing something here? $\endgroup$ – James Feb 26 at 11:10
  • $\begingroup$ @James I guess that is correct, in a way, but I was looking for a more specific criterion. For example, the identity matrix is equivalent to all matrices of the form $M M^T$, i.e. all symmetric matrices and in particular all diagonal matrices, but I personally wouldn't say that all diagonal matrices represent the same linear map (that I would usually refer to through the relation $A \sim M A M^{-1}$). Hope that makes sense. $\endgroup$ – Sebastian Schulz Feb 26 at 13:30
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    $\begingroup$ Mea culpa, I apologize. $\endgroup$ – James Feb 26 at 14:10
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    $\begingroup$ Is the work on "unitary congruence" relevant? Here's one random paper that came up in a search arxiv.org/abs/0710.1530 and there seem to be many others. These papers all allow "M" to be any unitary matrix though, not just those with det M = 1. $\endgroup$ – j.c. Feb 26 at 17:42
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    $\begingroup$ @j.c. YES! Thank you! The paper you shared didn't immediately help but the phrasing "unitary congruence" was enough to find the path to an answer here: link.springer.com/article/10.1007%2Fs10958-012-0780-9 $\endgroup$ – Sebastian Schulz Feb 26 at 19:33

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