The following exercise is taken from Permutation Groups by Peter Cameron.
Let $G$ be a primitive group of odd order. By the Feit-Thompson Theorem, $G$ is soluble, and so a minimal normal subgroup $N$ of $G$ is abelian. Show that $N$ acts regularly. Identify $\Omega$ with $N$ as in Section 1.7. Show that the permutation of $N$ given by $n \mapsto n^{-1}$ normalises $G$.
The identification he is refering to works by using that $N$ acts regular, hence for some fixed $\alpha$ we have a correspondence between the elements $\beta \in \Omega$ with the unique $n \in N$ such that $\beta = \alpha^n$.
That $N$ acts regularly follows easily, for as $\alpha^{Ng} = (\alpha^g)^N$ the $N$-orbits form a system of blocks, or by noting that $G_{\alpha}$ is maximal and we must have $G = G_{\alpha} N$ as $N \le G_{\alpha}$ could not be because the action is faithful.
So as I understand the exercise the inversion map is also a (self-inverse) permutation of $N$ (or $\Omega$ by the identification of both). So it asks to show that this permutation normalises $G$ in $Sym(\Omega)$, i.e. $\varphi \in N_{\operatorname{Sym}(\Omega)}(G)$ with $\varphi : N \to N$ being the inversion map. But this means that $((n^{-1})^g)^{-1}$ equals some permutation in $G$, but as conjugation acts as automorphisms on $N$ we have $$ ((n^{-1})^g)^{-1} = ((n^{-1})^{-1})^g = n^g. $$ So this is quite a trivial property, and it always holds if we have some regular normal subgroup (which could also be non-abelian). But the assumptions seem to be more specific, and also my solution seems really simple.
So have I understood anything wrong? The complicated assumptions and the fact that my argument is essentially very trivial make me somehow suspicious about it...