Does there exist 2 matricies, such that they can be used to transpose any n by n matrix? Ideally $\exists A, B$ to be able to transpose matrix $X \; \forall X \in M_{n\times n} $ by matrix multiplication. (Even more ideal is if there is only one matrix, $A$ that can transposes $X$ as follows: $X^T = AX$ but I'm ignoring this posibility for now)
So far, I'm guessing that such a matrix doesn't exist and have been trying to prove that. I've tried to prove $\not \exists A,B : \forall X\in M_{n\times n}, \: X^T = AXB $
I've examined a sub-case where $X$ is an orthogonal matrix, but that hasn't gotten me anywhere except $XAXB = I_n$ 

Presumably $n>1$. If $X^T=AXB$ for every $n\times n$ matrix $X$, then $vu^T=(Au)(v^TB)$ for every pair of nonzero vectors $u$ and $v$, meaning that for any fixed $u\ne0$, $Au$ is a nonzero multiple of any $v$. This is impossible because $n>1$.
