@OfirGorodetsky Thanks for the reference giving a description of $\mathbb{E}$. I was curious what that field would be.

@KConrad Thanks for pointing that out about the $\ell = p$ case. I had mostly been thinking of the $\ell \neq p$ case where, of course, it might happen that $\zeta \in \mathbb{Q}_{\ell}$ when $p \mid \ell -1$. That's not quite what I was expecting in the $\ell = p$ case. Perhaps I've not translated my question from rep theory correctly in the $\ell$-adic case.

@DavidLoeffler Ah, thanks, I was hoping to get some argument like that by passing to the smaller extension $\mathbb{E}$, so as to mimic the situation with $\mathbb{R}(\zeta)/\mathbb{R}$. I see now if you take $\mathbb{E} = \mathbb{Q}(\zeta+\zeta^{-1}) \subseteq \mathbb{K}$ to be the maximal real subfield then for any $x \in \mathbb{E}$ you have $\mathrm{N}_{\mathbb{K/F}}(x) = \mathrm{N}_{\mathbb{E/F}}(x)^2$ so it will be positive.

@DanielLitt "Still, it's a bit surprising to me that such an innocent statement should be so difficult." - Possibly a definition for groups.

Assume a module $M$ is determined up to isomorphism by its character $\chi$. Write $\chi = \chi_1 + \cdots + \chi_n$ as a sum of irreducibles. Let $M_i$ be a simple module affording $\chi_i$. The direct sum $M_1 \oplus \cdots \oplus M_n$ affords $\chi$ and by assumption is isomorphic to $M$. Hence $M$ is a direct sum of simple modules, so semisimple. Complete reducibility holds.

A Chevalley group over $\mathbb{C}$, like $\mathrm{SL}_n(\mathbb{C})$, is always perfect, i.e., equals its derived subgroup. So this forces no non-trivial characters because any commutator is in the kernel of a character, no?

You still need to add the normal assumption to the definition. Reductive groups do have non-trivial connected abelian unipotent subgroups. The root subgroups are such. They're just not normal.

In any case, you should look at the book "Unipotent and nilpotent classes in simple algebraic groups and Lie algebras" by Liebeck and Seitz. It has all the information you will need.

As you say, you just need to check whether a class is invariant under the automorphism induced by the Frobenius. In good characteristic it is enough to check that the weighted Dynkin diagram of the class is invariant under the automorphism. If $G$ is simple then this is true unless $G$ is type $\mathsf{D}_{2n}$. In this case there are a family of classes which are interchanged by the automorphism.

@LSpice Urgh, thanks for checking my stupidity. The comment is so stupid I'm just going to delete it.

This is false for a general connected reductive algebraic group but is true if the group is simple and simply connected. This is discussed in Steinberg's "Endomorphisms of linear algebraic groups". Parametrising the simple modules in the general case is disussed by Brunat and Lübeck in the following paper arxiv.org/pdf/1211.3692.pdf.