Let $G$ be a finite group, and let $\sigma: G \to GL\_n(k)$ be a (not necessarily irreducible) representation defined over an algebraically closed field $k$ of characteristic $p$. Let $\sigma^{(m)}$ denote the mth frobenius twist of $\sigma$, that is $\sigma\_{ij}^{(m)}(g) = (\sigma\_{ij}(g))^{p^m}$. Note that if $\sigma^{(m)} = \sigma$ then the image of $\sigma$ lands inside $GL\_n(\mathbb{F}\_{p^m})$.


Here are two notions:


1. If $\sigma$ is isomorphic to (i.e. conjugate to) a representation $\tilde{\sigma}$ such that $\tilde{\sigma}^{(m)} = \tilde{\sigma}$, then I would say "$\sigma$ is defined over $\mathbb{F}\_{p^m}$".
2. If instead we only know that $\sigma^{(m)}$ and $\sigma$ are isomorphic, it seems natural to say that "the isomorphism class of $\sigma$ is defined over $\mathbb{F}\_{p^m}$".


My first question is: Are these actually different notions? My gut says yes, but I couldn't come up with a counterexample this afternoon.


If the answer is yes, my follow up is: Is there a reference for working with this second notion of the field of definition of an isomorphism class of representations?



Field of definition for isomorphism classes of modular representations
The two notions are the same.


Clearly the first implies the second.


Assume that $\sigma^{(m)}$ is isomorphic to $\sigma$. So there is some $a\in GL\_n(k)$ such that $a\sigma^{(m)}(g)a^{-1}=\sigma(g)$ for all $g\in G$.


By [Lang's Theorem](https://en.wikipedia.org/wiki/Lang%27s_theorem), there is some $b\in GL\_n(k)$ with $a=b^{-1}b^{(m)}$. Then $b^{(m)}\sigma^{(m)}(g)(b^{(m)})^{-1}=b\sigma(g)b^{-1}$. So the representation $\tilde{\sigma}=b\sigma b^{-1}$, which is conjugate to $\sigma$, satisfies $\tilde{\sigma}^{(m)}=\tilde{\sigma}$.


