TITLE: Conjugate subgroups of $GL_n(K)$
QUESTION [6 upvotes]: Let $ K \subset L$ two fields,  $G$ and $G'$ subgroups of $\mathrm{GL}_n(K)$. 
  Assume that $G$ and $G'$ are conjugate in $\mathrm{GL}_n(L)$. 

Are $G$ and  $G'$ conjugate in  $\mathrm{GL}_n(K)$?
I have a solution for this one using a non trivial paper. I will add the source later because I would like to know if someone have a simpler proof in that case.

REPLY [7 votes]: It sounds as though you know that the answer to this question is yes.
I think it is essentially the same result as the following, which is Theorem 29.7 of the standard textbook on Representation Theory by Curtis and Reiner.
If two representations of a finite group over a field $K$ are equivalent when regarded as representations over a field $L$ containing $K$, then they are equivalent as representations over $K$.
This result seems to be less well known than might be expected. You would expect it to be natural question, for example about complex representations that can be written over ${\mathbb R}$.