$G$ is a group of odd order, $\sigma$ is an automorphism of $G$, and $\sigma^2=\mathrm{id}$.


I want to find an example to show that
$G\_s= \{ g \in G \mid \sigma \left( g \right)= g^{-1} \} $ might not be a subgroup of $G$.


$G$ has to be a non-Abelian group.
If $G\_s$ is a group, it should be an Abelian group.
I tried some examples (the orders of most examples are not very large), but I didn't find counterexamples.



Find an example where a subset of “inverse fixed points“ is not a subgroup
For an odd prime $p$, let $$G = \langle a,b \mid a^p=b^p=[[a,b],a]=[[a,b],b]=1 \rangle$$ be an extraspecial group of order $p^3$ and exponent $p$.


Then there is an automorphism $\sigma$ of $G$ with $\sigma^2=1$, $\sigma(a)=a^{-1}$, $\sigma(b)=b^{-1}$, and $\sigma([a,b])=[a,b]$.


Since $a,b \in G\_s$ but $G\_s \ne G$, it is clear that $G\_s$ cannot be a subgroup of $G$. In fact $|G\_s| = p^2$.


