Quotient-pullbackable implies linearly pushforwardable for representation over prime field

Statement

Suppose $\sigma$ is an automorphism of a group $G$. Suppose $\rho:G \to GL(V)$ is a finite-dimensional linear representation of $G$ over a prime field (i.e., either the field is of prime order or is the field of rational numbers). Consider the corresponding semidirect product $H = V \rtimes G$. Suppose, further, that there exists an automorphism $\sigma'$ of $H$ such that $\sigma'$ restricts to an automorphism $\alpha$ of $V$, and the automorphism induced on $G$ as a quotient equals $\sigma$. Then, we have: $\rho \circ \sigma = c_\alpha \circ \rho$

where $c_\alpha$ denotes conjugation by $\alpha$.

Note that we need the field to be a prime field in order to ensure that any automorphism of the vector space as an abelian group is also a linear automorphism.

Facts used

1. Automorphism group equals general linear group for vector space over prime field
2. Automorphism group action lemma for quotients: Suppose $A$ is an Abelian normal subgroup (?) of a group $H$, and $\sigma'$ is an automorphism of $H$ that restricts to an automorphism $\alpha$ of $A$. Note that this also shows that $\sigma'$ descends to an automorphism, say $\sigma$, of $H/A$.

Since $A$ is abelian, we have an action of the quotient group on it by conjugation (see quotient group acts on Abelian normal subgroup), giving a homomorphism: $\rho:H/A \to \operatorname{Aut}(A)$.

The claim is that: $\rho \circ \sigma = c_\alpha \circ \rho$.

where $c_\alpha$ denotes conjugation by $\alpha$ as an element of $\operatorname{Aut}(A)$.

Proof

Given: A group $G$, an automorphism $\sigma$ of $G$, a representation $\rho:G \to GL(V)$ for a vector space $V$ over a prime field. There is an automorphism $\sigma'$ of $H = V \rtimes G$ such that $\sigma'$ restricts to an automorphism $\alpha$ of $V$, and it induces the automorphism $\sigma$ on the quotient $G$.

To prove: $\rho \circ \sigma = c_\alpha \circ \rho$.

Proof: Since $F$ is a prime field, $GL(V)$ is the whole automorphism group of $V$ by fact (1) (in general, it is a proper subgroup). Thus, the element $\alpha$, which is a group automorphism of $V$, is actually in $GL(V)$. Thus, fact (2), setting $H = H, A = V, \sigma' = \sigma', \alpha = \alpha, \sigma = \sigma$, gives the desired result.