Let $E/F$ be a field extension and $n$ a positive integer. Given two ordered subsets $U=\{A\_1,\dots,A\_k\}$ and $V=\{B\_1,\dots,B\_k\}$ of $M\_n(F)$. If there is a matrix $P\in GL\_n(E)$ such that $PA\_iP^{-1}=B\_i$ for every $1\le i\le k$, then can we find an element $P'\in GL\_n(F)$ with $P'A\_iP'^{-1}=B\_i$ for every $i$.

The simplest case with $k=1$ is well-known.

An evidence is as follows. Assume that $F$ is of characteristic $0$. Let $G\_1,G\_2$ be two finite subgroups of $GL\_n(F)$. If there is $P\in GL\_n(E)$ such that $PG\_1P^{-1}=G\_2$, then there is $P'\in GL\_n(F)$ such that $P'G\_1P'^{-1}=G\_2$.

A proof: Let $\rho\_1:G\_1\to GL\_n(F)$ be the injection, and $\rho\_2:G\_1\to GL\_n(F)$ is defined by $\rho\_2(h)=PhP^{-1}$. Each $\rho\_i$ is a completely reducible representation of the group $G\_1$ by Maschke's theorem. Moreover, for every $h\in G\_1$, the traces coincide $Tr(\rho\_1(h))=Tr(\rho\_2(h))$. By Bourbaki lemma (Corollary 3.8 p.650 of Algebra (3rd edition) by S.Lang),
the two representations are isomorphic. That means there is $P'\in GL\_n(F)$ such that $PhP^{-1}=P'hP'^{-1}$ for every $h\in G\_1$. In particular, $P'G\_1P'^{-1}=G\_2$.

Any comments are welcome.