Does there exist $g\in Aut(G)$, but $g^2\neq e$?

Question

Let $G(V,E)$ be a self-complementary graph such that $|V|\geq 2$
Does there exist $g\in \text{Aut}(G)$, but $g^2\neq e$ ?

All i know is that :

Given that $G$ is self-complementary, we have $G \cong \overline{G}$, where $\overline{G}$ denotes the complement of $G$. Thus, every automorphism of $G$ is also an automorphism of $\overline{G}$, and vice versa.

Now, let's consider an automorphism $g \in \text{Aut}(G)$. Since $g$ is an automorphism of $G$, it maps vertices to vertices.

Let $v \in V(G)$. Since $G$ is self-complementary, there exists a vertex $u$ in $G$ such that $u$ and $v$ are not adjacent (i.e., ${u, v} \notin E(G)$) and $u \neq v$.

Answer 1

If $$ is the graph of the $5$-cycle, then $$ has an automorphism of order $5$. (A rotation.)