What is $\text{Out}(G) = \text{Aut}(G) / \text{Inn}(G)$ when $G = \mathbb{Z}/n\mathbb{Z}^\times$? For example, when $p$ is prime, which automorphisms of $\mathbb{Z}/p\mathbb{Z}^\times \cong \mathbb{Z}/(p-1)\mathbb{Z}^+$ arise as conjugation, and what happens when we quotient $\text{Aut}(\mathbb{Z}/p\mathbb{Z}^\times) \cong \mathbb{Z}/(p-1)\mathbb{Z}^\times$ by $\text{Inn}(\mathbb{Z}/p\mathbb{Z}^\times)$?
EDIT: $\mathbb{Z}/n\mathbb{Z}^\times$ is abelian... I'm posting the question because I hope it will save time for other people in the future.
While writing the question I found A question about automorphism and inner automorphism of $\mathbb{Z}$ and realized there is a very simple answer.
$\mathbb{Z}/n\mathbb{Z}^\times$ is abelian since $\mathbb Z$ is abelian. And if $G$ is an abelian group, then $\text{Inn}(G) = \{ x \mapsto gxg^{-1} : g \in G \} = \{ x \mapsto x \}$, so $\text{Inn}(G)$ is trivial. Another way to see this is that $Z(G) = G$, so $\text{Inn}(G) \cong G / Z(G) = G/G \cong \{ 1 \}$. Thus $\text{Out}(G) = \text{Aut}(G)/\{\text{Id}_G\} \cong \text{Aut}(G)$.
So answering Find all automorphisms of the multiplicative group mod $n$ would answer the unnecessarily complicated question I asked.