Ytukyg

To show that,

$operatorname{Aut}(mathbb{Z}/nmathbb{Z})cong (mathbb{Z}/nmathbb{Z})^times.$

What I feel is that if we show that "any automorphism $phi:mathbb{Z}/nmathbb{Z}longrightarrowmathbb{Z}/nmathbb{Z}$ is of the form $phi_a([k]_n):=[ak]_n$ for some $ainmathbb{Z}$ such that gcd$(a,n)=1.$ then we are done.

But I am not sure how to do this?
Thanks in advance.

$mathbb{Z}/nmathbb{Z}$ is cyclic generated by $1$. So an isomorphism $phi$ is completely determined by what 1 gets map to. Because suppose $phi(1)=k$ then $phi(m)=phi(1+1+1+...+1)=phi(1)+phi(1)+...+phi(1)= m*k$. Also, you need to make sure $0$ is the only element that gets map to $0$ as we have a isomorphism not a homomorphism. So you can't have $f(1)=k$ where $k$ is a zero divisor as then f(z)=zk=0 for some $z neq 0$. Which means $f(1)$ has to be invertible. Now you can check that $alpha :Aut(mathbb{Z}/nmathbb{Z}) rightarrow (mathbb{Z}/nmathbb{Z})^*$ by $alpha(f)=f(1)$ is an isomorphism.

Automorphism ring is a way to construct ring from groups. you can check that the reason why multiplication of two negative numbers is positive is because the ring of integers is actually automorphism of the abelian additive group $mathbb{Z}$. Because when you compose $f,g$ where $f(1)=-k$(sends all positive to negative),$g(1)=-n$(sends all positive to negative and vice-versa), you get that the composition of the two map sends all positive to all positive. Hope I got it right.