I have posted this elsewhere and got only a partial reply. I don't know whether this qualifies the question for an open-problem tag; if it does, please anyone insert it.

Let $L$ be a field, and $K$ a subfield of $L$. Let $n$ and $m$ be two nonnegative integers.

For any $n\times n$ matrix $A \in \operatorname{M}_n\left(K\right)$ and any $m\times m$-matrix $B \in \operatorname{M}_m\left(K\right)$, and any field $S$ containing $K$, we define \begin{align} \rho_{S}\left( A,B\right) := \max\left\{\operatorname{Rank} Q \mid Q\in\operatorname{M}_{n,m}\left( S\right) ; \ AQ = QB\right\}. \end{align} We can call this number $\rho_{S}\left( A,B\right)$ the "conjugacy rank" of the matrices $A$ and $B$ over the field $S$.

(Note that if $n = m$, then this conjugacy rank is directly connected with conjugacy -- i.e., similarity -- of matrices: Namely, in this case, we have $\rho_{S}\left( A,B\right) = n$ if and only if the matrices $A$ and $B$ are conjugate to each other in the ring $\operatorname{M}_{n}\left( S\right)$.)

My question is: Do we have $\rho_{K}\left( A,B\right) = \rho_{L}\left( A,B\right)$ for any two matrices $A \in \operatorname{M}_n\left(K\right)$ and $B \in \operatorname{M}_m\left(K\right)$ ?

This can be shown in the case of $n = m \leq \left\vert K\right\vert$ by a "polynomials which vanish everywhere must be identically $0$" argument. Besides, in the case of $\rho_{L}\left( A,B\right) = n = m$, it can be shown using the rational canonical form. I am interested in the most general case of the problem -- neither restricting $\left\vert K\right\vert$, nor $\rho_{L}\left( A,B\right)$, nor requiring $n = m$.

What also might be of help: For any field $S$ containing $K$, the space
$$R_{S}\left( A,B\right) = \left\{ Q\in\operatorname{M}_{n,m}\left( S\right) \mid AQ = QB\right\}$$
is a subspace of the vector space $\operatorname{M}_{n,m}\left(S\right)$.
Besides, every basis of the space $R_{K}\left( A,B\right)$ is also a basis of the space $R_{S}\left( A,B\right)$ for every field $S$ containing $K$. However, *this alone is not enough*; you can easily construct a subspace of $\operatorname{M}_{n}\left(\mathbb{F}_p\right)$ that consists of singular matrices only but loses this property when extended into $\operatorname{M}_{n}\left(\mathbb{F}_{p^2}\right)$.