Conjugate subgroups by permutation matrices Is the following statement true or false?
In $G=\operatorname{GL}_{n}(\textbf{C})$, two elementary abelian $2$-subgroups $X$ and $Y$ of $G$ whose generators are all diagonal matrices are conjugate if and only if there is a permutation matrix $P$ s.t. $PX=YP$.
I'm aware that the right implies the left. For the other direction, it is true when $X$ and $Y$ has only one generator because the numbers of $-1$ on the diagonals of the generators are equal. What about when we have more than one generator? Are there cases where $X,Y$ can only be conjugate by a non-permutation matrix? I checked some examples and didn't see such a case. Thank you.
 A: Suppose the characters of an abelian group $A$ are $\chi_1,\cdots,\chi_k$. Then any $A$-rep $V$ has an isotypical decomposition $V=V_1\oplus\cdots\oplus V_k$
where each $V_i$ is a sum of irreps corresponding to $\chi_i$. The components are weight spaces, i.e. $V_i=\{v\in V\mid av=\chi_i(a)v~~\forall a\in A\}$. If $V=\mathbb{C}^n$ and $A$ is diagonal, in fact
$$ V_i=\mathrm{span}\{e_j\mid ae_j=\chi_i(a)e_j~~\forall a\in A\}, $$
where $e_1,\cdots,e_n$ are the basis coordinate vectors. To this, we can associate a partition $\Pi$ of $\{1,\cdots,n\}$, where the $i$th block consists of all $j$ for which $e_j\in V_i$.
A conjugate representation ${}^g\!A=gAg^{-1}$ of $A\,$ ($g\in\mathrm{GL}_n\mathbb{C}$) has isotypic decomposition $gV_1\oplus\cdots\oplus gV_k$, with $\chi_i$ associated to each $gV_i$ again. Thus, the component dimensions are unchanged, which means if ${}^g\!A$ is also diagonal then it too has an associated partition $\Pi'$ and of the same shape (i.e. block sizes) as $\Pi$.
Since $S_n$ acts transitively on ordered partitions of the same shape, there is a permutation $\sigma$ for which $\Pi'=P\Pi$, and if $P$ is the associated permutation matrix then ${}^g\!A={}^P\!A$ because both groups act the same on all standard basis vectors.
