Simultaneously conjugate finitely matrices 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.
 A: Consider the monoid $G$ generated by $A_1,  \dots, A_n$. We have two modules over the monoid algebra $F[G]$ just like in your solution to a special case. These modules are isomorphic after we base change to $E$. As they are also finite-dimensional, we can conclude that they are isomorphic by the Noether-Deuring theorem (which holds for modules over an algebra and not just for group representations). This implies what we want.
Here's the statement of Noether-Deuring that I'm using: let $A$ be an algebra over a field $F$ and let $M,N$ be modules over $A$ which are finite-dimensional over $F$. Let $E/F$ be any field extension. Then if $E\otimes_F M\cong E \otimes_F N$ as $E \otimes_F A$-modules, we have $M\cong N$.
A: Yes, if $F$ is infinite.
The set of equations $MA_i=B_iM,i=1,...,k$ is a linear system $CX=0$, where $C$ has entries in $F$ and $X=\pmatrix{m_{11} \cr m_{12}\cr \vdots \cr m_{nn}}\in F^{n^2}$.
Let $V=\{X\in F^{n^2}\mid CX=0\}$ and $V_E=\{X\in E^{n^2}\mid CX=0\}$
Note that an $F$-basis of $V$ is an $E$-basis of $V_E$. Indeed, if $(X_1,\ldots,X_d)$ is an $F$-basis of $V$, the matrix whose columns are $X_1,\ldots,X_d$ has rank $d$. But its also has rank $d$ when viewed as a matrix with coefficients in $E$. Hence $(X_1,\ldots,X_d)$ are also $E$-linearly independent. Now similarly, the rank of $C$ is the same , viewed has a matrix with entries in $F$ or in $E$. Consequently, $\dim_F(V)=\dim_E(V_E)$ and it follows that  $(X_1,\ldots,X_d)$ is an $E$-basis of $V_E$.
So let $(X_1,...,X_d)$ be an $F$-basis of $V$, and let $f=\det(T_1X_1+\cdots+T_d X_d)\in F[T_1,...,T_d]$.
Since $P\in V_E$, $P=t_1X_1+...+t_dX_d, t_j\in E$ by the previous point.  Since $P$ is invertible, $f(t_1,...,t_d)\neq 0$. Hence, $f$ is a nonzero polynomial. Since $F$ is infinite, it cannot be identically zero on $F^d$, meaning that there is an $F$-linear combination of $X_1,...,X_d$  which is invertible. This exactly means that our set of equations has an invertible solution with coefficients in $F$.
