Must normalizing field outer automorphisms "divide" the dimension?

Question

Imprecise question: To get a normalizing field outer automorphism of order $r$, must we multiply the dimension by $r$?

Precise hypothesis: Let $p\geqslant 5$ be a prime, let $q$ be a power of $p$ and let $H$ be an absolutely irreducible quasisimple subgroup of $G:=\textrm{GL}(n,q)$. Let $T:=H/\textrm{Z}(H)$ and suppose that $T$ is a simple group of Lie type defined over a field of characteristic $p$. Let $N$ be the normalizer of $H$ in $G$, and let $Z$ be the centre of $G$. By Schur's lemma, $\textrm{Z}(H)=Z\cap H$ has order dividing $q-1$.

Precise question: Does the $p$-part of $|N/H Z|$ divide the $p$-part of $n$?

Remark: There is a homomorphism $\phi\colon N\to\textrm{Out}(T)$ since $$N\to\textrm{Aut}(H)\to\textrm{Aut}(H/\textrm{Z}(H))=\textrm{Aut}(T)\to\textrm{Out}(T).$$ Furthermore $H$ and $Z$ (and hence $HZ$) are contained in the kernel of $\phi$. The induced homomorphism $N/HZ\to\textrm{Out}(T)$, when restricted to a Sylow $p$-subgroup of $N/HZ$, is injective because an automorphism of $H$ that acts trivially on $H/\textrm{Z}(H)$ has order dividing $|\textrm{Z}(H)|$ and hence is coprime to $p$. Thus $|N/HZ|_p$ divides $|\textrm{Out}(T)|_p$. Here the $p$-part of a positive integer $n$ (viz. the largest power of $p$ that divides $n$) is denoted $n_p$.

Note that $|\textrm{Out}(T)|=dfg$ where $d$, $f$, $g$ denote the number of "diagonal", "field" and "graph" outer automorphisms, respectively. We know $g_p=1$ since $p\geqslant 5$, and $d_p=1$ since $T$ has characteristic $p$. Hence $|\textrm{Out}(T)|_p=d_pf_pg_p=f_p$. Thus $|N/HZ|_p$ divides $f_p$. Does it divide $n_p$?

[Edit: Many thanks Derek. The previous example was not absolutely irreducible.]

Example: Let $V=({\mathbb F}_{p^{20}})^m$ and embed $H:=\textrm{SL}(V)$ into $G:=\textrm{GL}(\Lambda^r(V))$ where $\Lambda^r(V)$ is the $r$th exterior power of $V$. If the normalizer of $H$ in $G$ involves a field outer automorphism of order $5$, then must $\dim(\Lambda^r(V))=\binom{m}{r}$ be divisible by 5?

Answer 1

The answer to this question is No. Let $U$ be the natural module for $H=\textrm{SL}(2,5^5)$ and let $V=U\otimes U^\sigma\otimes\cdots\otimes U^{\sigma^4}$ where $\sigma$ is field automorphism $\lambda\mapsto\lambda^{5}$. Then $H$ is an absolutely irreducible subgroup of $\textrm{GL}(2^5,5^5)$ moreover $u\otimes u^\sigma\otimes\cdots\otimes u^{\sigma^4}\mapsto u^\sigma\otimes (u^\sigma)^\sigma\otimes\cdots\otimes (u^{\sigma})^{\sigma^4}$ normalizes $H$ and so $|N/HZ|_5$ is divisible by 5 but $\dim(V)=2^5$ is not. (I will ask the `right' question later and add a link to it.)