For matrices, under what conditions can we write $AB = BC$?

To the OP: do you really think that $AB=BA^T$ when $B$ is symmetric ?

The answer to your question:

find $X\in M_n$ s.t. $A B=BX$ (where $A\in M_m,B\in M_{m,n}$ are given)

can be written, using the Moore Penrose inverse. cf.

https://en.wikipedia.org/wiki/Moore%E2%80%93Penrose_inverse

If $BB^+AB\not= AB$, then no solutions.

If $BB^+AB=AB$, then the solutions are

$X=B^+AB+(I_n-B^+B)W$ where $W$ is an arbitrary $n\times n$ matrix.

If $B$ is full rank matrix then we have $C=B^{-1}AB$. Obviously then $C$ is similar to $A$ with all consequences of this.