TITLE: On $XX'=I$ such that $AX=XB$ is true
QUESTION [6 upvotes]: Given list of symmetric matrices $\{A_i,B_i\}_{i=1}^r\in\Bbb R^{n\times n}$ where $r\in\Bbb N$ is arbitrary what is a good description of collection of $X\in\Bbb R^{n\times n}$ such that $$XX'=I$$ $$A_iX=XB_i$$ holds?
Note: $X'$ means the transpose of $X$.
(1) Is there a test to see if there is no such $X$?
(2) Is it easy to see if we find one such $X$ we can find ALL such $X$ from finding $U\in\Bbb R^{n\times n} $ such that $UU'=I$ and $UA_i=A_iU$ or $B_iU=UB_i$ at every $i\in\{1,\dots,r\}$?
(3) What if we have $\Bbb F_{p^r}$ instead of $\Bbb R$?

REPLY [3 votes]: A necessary condition is that, for every word $w(x,y)$ in two letters, we have
$${\rm Tr}w(A_i',A_i)={\rm Tr}(B_i',B_i).$$
If $r=1$ (one pair only), this condition is also sufficient; this is Specht's Theorem. Notice that the symmetry assumption is not important here.
In the present case, a necessary condition is that for every word $w$ in $2r$ letters, one has
$$w(A_1,A_1',\ldots,A_r,A_r')=w(B_1,B_1',\ldots,B_r,B_r').$$
Whether it is also a sufficient condition is unclear to me.