If a tensor product of matrices is symmetric, are the underlying matrices also symmetric? Let $A_{i\tilde i} \in \mathbb{R}^{d^2},B_{j\tilde j} \in \mathbb{R}^{n^2}$ be real invertible square matrices.
Define $$ C_{ij,\tilde i\tilde j}:=A_{i\tilde i} B_{j\tilde j}$$ and suppose that $C_{ij,\tilde i\tilde j}=C_{\tilde i\tilde j,ij}$, that is
$$ A_{i\tilde i} B_{j\tilde j}=A_{\tilde i i} B_{\tilde j j},$$
for all $i,\tilde i,j,\tilde j$. 
Is it true that $A,B$ are symmetric, that is $A_{\tilde i i}=A_{i\tilde i}, B_{j\tilde j}=B_{\tilde j j}$
(More abstractly, the question is the following: suppose a tensor product of two non-degenerate bilinear forms is symmetric. Is it true that the factros are symmetric?)
 A: Partial answer: If $B$ has a diagonal element which is non-zero, say $B_{1,1}$, you can look at $A_{i\tilde i} B_{1, 1}$, and vice versa. More generally, if there is a single entry in one matrix that is equal to the transposed entry, while being non-zero, the other matrix must be symmetric. In general, however, I have no idea.
A: Arthur has given half the answer, so I'll give the other half.
$$\newcommand{\bmat}{\begin{pmatrix}}\newcommand{\emat}{\end{pmatrix}}
\bmat 0&-1\\1&0 \emat
\otimes
\bmat 0&-1 \\1 & 0 \emat
=
\bmat 0 & 0 & 0 & 1 \\
0 & 0 & -1 & 0 \\
0 &  -1 & 0 & 0 \\
1 & 0 & 0 & 0\emat,
$$
which is symmetric.
The idea is that when $A_{\tilde{i}i}$ and $B_{j\tilde{j}}$ aren't zero, the equation
$$A_{i\tilde{i}}B_{j\tilde{j}}=A_{\tilde{i}i}B_{\tilde{j}j}$$
can be rearranged to give
$$\frac{A_{i\tilde{i}}}{A_{\tilde{i}i}}=\frac{B_{\tilde{j}j}}{B_{j\tilde{j}}},$$
then since we can vary $j$ and $\tilde{j}$ while fixing $\tilde{i}$ and $i$, and similarly for $i$ and $\tilde{i}$, there is some constant $C$ such that (relabeling) when $A_{i\tilde{i}}\ne 0$, $\frac{A_{\tilde{i}i}}{A_{i\tilde{i}}}=C$, and when $B_{j\tilde{j}}\ne 0$, $\frac{B_{\tilde{j}j}}{B_{j\tilde{j}}}=C$.
Now if $C\ne 0$, then when $B_{j\tilde{j}}\ne 0$, $B_{\tilde{j}j}\ne 0$, so we get $C^{-1}=C$. Hence $C=\pm 1$. If $C=1$, the matrix is symmetric.
If $C=-1$ the matrix is antisymmetric. The point though is that the case $C=-1$, the case where both matrices are antisymmetric is a valid solution. I'm not entirely sure what matrices with $C=0$ look like, but based on the simplest such matrix, $\bmat 0 & 1 \\ 0 & 0 \emat$, it looks like they don't give valid solutions. Probably they're not invertible for one thing.
So I guess, if you formalize my work here that I used to construct a counterexample, you can probably at least salvage: If $A\otimes B$ is symmetric, then either $A$ and $B$ are both symmetric or both antisymmetric. Which is still nice.
A: This is just to write a complete answer, based on jgon's answer:
We shall prove that both $A,B$ are symmetric or both are anti-symmetric.
We start with the following observation: $A_{i\tilde i}=0 \Rightarrow A_{\tilde ii}=0$.
Indeed, $A_{i\tilde i}=0 \Rightarrow 0=A_{i\tilde i} B_{j\tilde j}=A_{\tilde i i} B_{\tilde j j}$ for all $j,\tilde j$. 
Since $B$ is invertible, there is some $B_{\tilde j j} \neq 0$, so $A_{\tilde ii}=0$.
Now, fix some $B_{j \tilde j } \neq 0$. Then, for every $i,\tilde i$ such that $A_{i\tilde i} \neq 0$, we have $ A_{\tilde ii} \neq 0$ and so by dividing the equation
$$ A_{i\tilde i} B_{j\tilde j}=A_{\tilde i i} B_{\tilde j j},$$ we obtain that
$$ \frac{A_{i\tilde{i}}}{A_{\tilde{i}i}}=\frac{B_{\tilde{j}j}}{B_{j\tilde{j}}} \, \, \text{is independent of } \, i,\tilde i, $$
So, whenever $A_{i\tilde i} \neq 0$, we have $$ \frac{A_{i\tilde{i}}}{A_{\tilde{i}i}}=C,$$ where $C \neq 0$ is a constant. Now, switching the roles of $i,\tilde i$ we deduce $C=C^{-1}$. Thus $C=\pm1$. 
This implies the equation $A_{i\tilde i}=CA_{\tilde i i}$ holds for all $i,\tilde i$. (The first observation implies this holds if $A_{i\tilde i}=0$, and we just proved this for all other cases).
This implies $A,B$ must be both symmetric or both anti-symmetric.
