My question is similar to How to find matrix $C$ satisfying $AB=CA$? but the matrix $A$ is rectangular. That is, $A \in \mathbb{R}^{m \times n}$, $m>n$ and $A$ has full column rank, and $B \in \mathbb{R}^{n \times n}$ is invertible.
Is there a matrix $C \in \mathbb{R}^{m \times m}$ such that $AB = CA$?
I wonder if the statement '$AB = CA$ if and only if $ker(A) \subseteq ker(AB)$' (from How to find matrix $C$ satisfying $AB=CA$?) is true for the case of rectangular, full column rank $A$.
I can see that $ker(A) \subseteq ker(AB)$ is necessary for the existence of $C$, but I'm not sure if it is a sufficient condition for the existence of $C$.
Edit: since $A$ has full column rank (i.e. linearly independent columns), we have $A^+ = (A^TA)^{-1}A^T$, and let $C = ABA^+$, then $CA = (AB(A^TA)^{-1}A^T)A = AB$.