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 $L$. For any two $n\times n$ matrices $A$ and $B$ from $\mathrm{M}_n\left(K\right)$, and any field $S$ containing $K$, we denote $$\rho_{S}\left( A,B\right) = \max\left\{\mathrm{Rank}Q\mid Q\in\mathrm{M}_{n}\left( S\right) ;\ AQ = QB\right\}.$$ We can call $\rho_{S}\left( A,B\right)$ the "conjugacy rank" of the matrices $A$ and $B$ over the field $S$ (noting that $\rho_{S}\left( A,B\right) = n$ if and only if the matrices $A$ and $B$ are conjugate to each other in $\mathrm{M}_{n}\left( S\right)$).

My question is: Do we have $\rho_{K}\left( A,B\right) = \rho_{L}\left( A,B\right)$ for any two matrices $A$ and $B$ from $\mathrm{M}_{n}\left( K\right)$ ?

This can be shown in the case of $n\leq\left\vert K\right\vert$ by a "polynomials which vanish everywhere must be identically $0$" argument. Besides, in the case of $\rho_{L}\left( A,B\right) = n$, 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 $\rho_{L}\left( A,B\right)$ -, but there may even be counterexamples.

What also might be of help: For any field $S$ containing $K$, the space
$$R_{S}\left( A,B\right) = \left\{ Q\in\mathrm{M}_{n}\left( S\right) \mid AQ = QB\right\}$$
is a subspace of the vector space $\mathrm{M}_{n}\left(S\right)$.
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 $\mathrm{M}_{n}\left(\mathbb{F}_p\right)$ that consists of singular matrices only but loses this property when extended into $\mathrm{M}_{n}\left(\mathbb{F}_{p^2}\right)$.