I'm stuck at this exercise. How can I approach this? I know that an isomorphism is a bijective homomorphism but we can't just count the number of bijections ($n!$), since we can't just assign every element to any other element. We need to work with the generators. A cyclic group of order n has $\phi (n) = |\mathbb Z ^{*}_{n}|$ generators. And this is where I get stuck. How can I continue from here?
Thank you
On
You are making a mistake on the number of generators. But anyway, notice that a morphism from a cyclic group $C_n$ of order $n$ is completely determined by the image of any one of its generators. So let $1\in C_n$ be a generator, where can you map it into the other cyclic group in such a way that you get a homomorphism of groups?
An homomorphism $\iota$ from $\mathbb{Z}_n$ onto itself is completely determined by the image of $1$, since $1$ generates $\mathbb{Z}_n$. Since you want $\iota$ to be surjective, $\iota(1)$ must be a generator of $\mathbb{Z}_n$ and, as you know, you have $\phi(n)$ choices for that. Therefore, the answer is $\phi(n)$.