Q:

Aut $\mathbb Z_p\simeq \mathbb Z_{p-1}$

I'm trying to prove that Aut $\mathbb Z_p\simeq \mathbb Z_{p-1}$ (p prime).
I know that Aut $\mathbb Z_p$ has $p-1$ elements because $\mathbb Z_p$ has $p-1$ possiblities of generators, so intuitively I see Aut $\mathbb Z_p\simeq \mathbb Z_{p-1}$, but I couldn't prove it formally.
I'm trying to build an isomorphic function, but I don't how to do it.
Am I in the right way? Maybe I'm forgetting some trick or something.
Thanks

A:

What you've said so far is correct, but the fact that the group is cyclic of order $p-1$ is significantly harder.  You'll need some real idea here; it's not a matter of just following your nose.
One standard approach is to prove the Cyclicity Criterion: a group $G$ of finite order $n$ is cyclic $\iff$ for each divisor $d$ of $n$ there are at most $d$ elements $x$ with $x^d = 1$ (the identity element).  This is generally proved using a counting argument, and it is helpful to know that $\sum_{d \mid n} \varphi(d) = n$, where $\varphi(d) = \# (\mathbb{Z}/d\mathbb{Z})^{\times}$ is the number of integers $1 \leq e \leq d$ with $\gcd(e,d) = 1$.  
With that in hand you should observe that $\mathbb{Z}/p\mathbb{Z}$ is a field, and thus the number of roots of the polynomial $x^d - 1 = 0$ in it is at most $d$.  This argument ends up showing a little more: any finite subgroup of the multiplicative group of a field is cyclic.  
P.S.: If I tell you that the result is often called "existence of primitive roots modulo $p$", that may help you look it up in many standard introductory texts on algebra and/or number theory.  

