If $\mathbb Z/2\mathbb Z \cong S_2$, why isn't $\mathrm{Aut}(\mathbb Z/2\mathbb Z) \cong \mathrm{Aut}(S_2)$?

Question

We know $\mathrm{Aut}(\mathbb Z/n\mathbb Z) \cong (\mathbb Z/n\mathbb Z)^\times$. So $\mathrm{Aut}(\mathbb Z/2\mathbb Z) \cong \{0\}$.

In Dummit & Foote, Section $4.4$, Proposition $17$ says if $n \ne 6$, then $\mathrm{Aut}(S_n) \cong S_n$. So we have $\mathrm{Aut}(S_2) \cong S_2$.

We know $\mathbb Z/2\mathbb Z \cong S_2$.

How come we don't have $\mathrm{Aut}(\mathbb Z/2\mathbb Z) \cong \mathrm{Aut}(S_2)$?

Answer 1

$\operatorname{Aut}(S_n) \cong S_n$ for $n \ne 2$ or $6$. Actually the rule is that every automorphism of $S_n$ is inner for $n \ne 6$ but because $S_2$ is Abelian, every inner automorphism is trivial.