Suppose that $\lvert G \rvert = 2$. Then I want to show that $\mathrm{Aut} (G) = \langle e \rangle$. I know that $G \cong \mathbb{Z}_2 \cong S_2 \cong D_2$ (by Cauchy's Theorem all of which are cyclic and abelian).
Let $\beta: G \to G$ be an automorphism. Then I know that $G$ is cyclic (let's say that $G$ is generated by $a$, i.e., $\langle a \rangle = \{a : a^2 = e\} = \{e, a\}= G$).
Is it true that $\beta$ must be the $\mathrm{id}$ map? And do I just need to prove that $\lvert \mathrm{Aut}(G) \rvert$ = 1?