$\operatorname{Aut}(\mathbb Z_n)$ is isomorphic to $U_n$. I've tried, but I can't solve the question. Please help me prove that:
$\operatorname{Aut}(\mathbb Z_n)$ is isomorphic to  $U_n$.

(If you know about ring theory.) Since $\mathbb Z_n$ is an abelian group, we can consider its endomorphism ring (where addition is component-wise and multiplication is given by composition). This endomorphism ring is simply $\mathbb Z_n$, since the endomorphism is completely determined by its action on a generator, and a generator can go to any element of $\mathbb Z_n$. Therefore, the automorphism group $\mathrm{Aut}(\mathbb Z_n)$ is the group of units in $\mathbb Z_n$, which is $U_n=U(\mathbb Z_n)$.
