Does the set of automorphisms of the integers modulo n have the same number of elements as the symmetric group of order n?

Question

As stated in the question header, does the set of automorphisms of the ring of integers modulo n have the same number of elements as the symmetric group of order n?

I tried creating the permutation (0,1) using the automorphisms in the ring $Z_n$, but couldn't do it successfully. I found that the group of automorphisms is restricted. Which is to say, no combination of multiplications and additions in $Z_n$ could produce (0,1), which is a necessity to generate all of $S_n$. (0,1,..,n-1) is of course present with +1. ($(12)$ and $(123\dots n)$ are generators of $S_n$).

Although I think I've found an answer, how can I prove this?

Thanks so much!

Answer 1

For every ring $R$ and every automorphism $f : R \to R$, we have $f(0)=0$. This follows from two facts: the additive identity of a ring is unique; and $f(0)$ is an additive identity of $R$.

So that already tells you that if $n$ is the number of elements of a finite ring $R$ then the number of automorphisms is $\le (n-1)!$ which, of course, is strictly less than $n!$.