Relevant Proposition
Suppose $G$ and $H$ are groups and that $e$ denotes the identity in each. If $\phi: G \to H$ is a group homomorphism then $\phi(e) = e$. Furthermore for all $g \in G$ and $n \in \mathbb{Z}$ we have $\phi(g)^n = \phi(g^n)$
By the proposition we know that $$\phi([1])^k = \phi([1]^k)$$ where $[1]^k = [1] + [1] + \cdots + [1] = [(k)( 1)] = [k]$. This means that the mapping of any element $[r] \in \mathbb{Z}_n$ by the automorphism $\phi$, $\phi([r])$, is determined by iterating the map of $[1]$ under $\phi$ $r$-times. Hence any automorphism $\phi \in Aut(\mathbb{Z}_n)$ is determined by $\phi([1])$.
For the next part, showing that $\phi([1])$ is a generator of $\mathbb{Z}_n$ I always seem to struggle with showing that an element is a generator of a cyclic group. I saw a proof of a similar problem and have interpreted it as follows:
By the proposition we know that $\phi([x]^k) = \phi([x])^k$ and $\phi([0]) = [0]$ for all $[x] \in \mathbb{Z}_n$ and $k \in \mathbb{Z}$. Therefore, if $\phi([1])^k = [0]$ for $k \lt n$ then by applying the inverse map $\phi^{-1}$ (whose existence is guaranteed by the fact that $\phi$ is an automorphism). We find: $$[1]^k = [0]$$ Now, first, is this the correct proof that $\phi([1])$ is a generator for $\mathbb{Z}_n$ and second, I'm having trouble seeing why this proves $\phi([1])$ is a generator. Is it because we see that given the proposition and the assumptions our algebra leads us to the fact that $[1]^k = [0]$ where we know that $[1]$ is a generator for the (additive) group $\mathbb{Z}_n$? Thus $\phi([1])$ must also be a generator?
Since every $k\in\Bbb Z_n$ is $1^k$ (multiplicative notation), a homomorphism of a cyclic group is completely determined by what it does to any generator.
And the homomorphism $\phi$ is bijective precisely when $\phi(1)$, say, is a generator.
A nice application is that, since the generators of $\Bbb Z_n$ are the $m$ such that $(m,n)=1$, we easily get that $\rm{Aut}(\Bbb Z_n)=(\Bbb Z_n)^×$, the group of units.
But, this is a little more advanced, because it uses the idea that $\Bbb Z_n$ is a ring.