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 $XAXB = I_n$ 

If we stack row by row the $n\times n$ matrices into $n^2\times 1$ vectors, then the OP's question is:
Is the function $f:X\in M_n\rightarrow X^T\in M_n$ (we can present $f$ in the form of a $n^2\times n^2$ matrix) decomposable into a tensor product $A\otimes B^T$ ? 
The answer is no and it's not a scoop ! Yet, $f$ can be written in a sum of $n^2$ such applications (perhaps, we can do better).
Proof. Let $E_{i,j}$  be the $n\times n$ matrix s.t. all the entries are $0$ except the $(i,j)$ one which is $1$. The matrix $E_{i,j}\otimes E_{k,l}$ is a $n^2\times n^2$ block-matrix s.t. all blocks are $0$ except the $i,j$ one which is $E_{k,l}$, that is,  this matrix has only one non-zero entry  (and it's $1$). Conversely, each matrix that has only one non-zero entry  (and it's $1$) is decomposable as above.
Note that $f$ is a permutation with square $I_{n^2}$, that is a product of disjoint transpositions. Thus $f$ has exactly $n^2$ non-zero entries, and, consequently, is the sum of $n^2$ such decomposable tensor products.
EDIT. Although my post does not interest anyone, I give below a result concerning decomposable permutations (the proof is not difficult).
$\textbf{Proposition}$. Among the $(n^2)!$ permutations of $ n ^ 2 $ elements, only those of the form $U\otimes V$, where $U,V$ are permutations of $n$ elements are decomposable (thus, only $(n!)^2$ permutations are decomposable).
