"Conjugacy rank" of two matrices over field extension

I think this is true, and can be proved by brute force: write an explicit formula for conjugacy rank. I'll prefer to restate the problem in terms of modules.

To an $n\times n$ matrix $A$ over a field $K$, associate the $K[x]$-module $M$ that is $K^n$ as a vector space, while $x$ acts as $A$. Everywhere below, all $K[x]$-modules are finite-dimensional as $K$-vector spaces. Then your definition becomes as follows:

Let $M$ and $N$ be two $K[x]$-modules. Define their conjugacy rank $\rho(M,N)$ to be the maximal dimension (over $K$) of a $K[x]$-module that is simultaneously isomorphic to a submodule of $M$ and a quotient-module of $N$. We aim to prove that $\rho(M,N)$ is stable under field extensions of $K$.

By structure theorem for modules over PID, we can write $M\simeq\bigoplus K[x]/f_i$, where invariant factors $f_i=f_i(M)\in K[x]$ satisfy $f_{i+1}|f_i$. It is easy to check the following claim:

Lemma: $M'$ is isomorphic to a quotient of $M$ if and only $f_i(M')|f_i(M)$. The same criterion holds for $M'$ being isomorphic to a submodule of $M$.

Corollary: There is unique up to isomorphism maximal-dimensional module $M'$ that is simultaneously isomorphic to a submodule of $M$ and a quotient-module of $N$; its invariant factors are given by $f_i(M')=gcd(f_i(M),f_i(N))$.

Since the formula for $M'$ is stable under field extensions of $K$, the claim follows.