I'm interested in characterizing the $m×n$ matrices $M$ such that for a fixed $m×m$ matrix $A$, there exists an $n×n$ matrix $B$ such that $M=AMB$.
I'm not sure if it makes the question easier or harder, but I'd particularly like to characterize the cases where $M$ is the incidence matrix of a graph.
I've been looking through matroid theory literature (due to the graphic matroid connection with incidence matrices), but doing column operations seems out of scope there.
Is there a neat way to approach this? Has this concept been studied at all? I'm not really sure what sort of characterization I can hope for here, so any potential connections would be appreciated!
Edit: In my case $A$ will always be non-singular. I'd also be interested if any characterization is known for particular (non-trivial!) examples of $A$. Furthermore, all of these matrices are over $\mathbf{C}$.
If $K$ and $L$ and $m\times n$ then there is an $n\times n$ matrix $B$ such that $K=LB$ if and only if $\mathrm{Im}(K)\subset\mathrm{Im}(L)$, that is, if each column of $K$ is a linear combination of the columns of $L$ (this is what the relation $K=LB$ says in the matrix form). If, in addition, both $K$ and $L$ have rank $n$, the matrix $B$ is unique and nonsingular. So the question is, for which $A$ (or $M$) the above holds with $K:=M$ and $L:=AM$.
The simplest thing to do in order to solve $M=AMB$ is to find a $B$ such that $\|M-AMB\|_F$ is minimal. This is essentially an ordinary least squares problem for columns of $B$ with the solution given by $$ B=(AM)^\dagger M, $$ where $(\cdot)^\dagger$ denotes the Moore-Penrose pseudoinverse. The residual matrix is $$ R:=M-AMB=M-AM(AM)^\dagger M=(I-AM(AM)^\dagger)M, $$ so the problem has a solution if and only if $R=0$. You can maybe use this to find a reasonable characterization.
A simple condition for solvability could hence be that $AM(AM)^{\dagger}=I$. If $M$ has full row rank, then $(AM)^\dagger=M^\dagger A^{-1}$ and hence $AM(AM)^{\dagger}=AMM^{\dagger}A^{-1}=I$ (if $M$ has full row rank, $M^\dagger$ is a right inverse of $M$). A weaker condition requires that $AM(AM)^{\dagger}M=M$ for which I don't know about any simple characterization except that mentioned in the first paragraph of my answer ($AM(AM)^{\dagger}$ is the orthogonal projector onto the image of $AM$ so $AM(AM)^{\dagger}M=M$ says simply that columns of $M$ are contained in the column space of $AM$).