The automorphism group of the group $G$, $\text{Aut}(G)$, is the group of isomorphisms from $G$ to $G$.
It is the set of functions $\phi$ that send the generators of $G$ to the generators of $G$, right?
I want to show that $\text{Aut}(\mathbb{Z}_6)$ is isomorphic to $\mathbb{Z}_2$.
Since the generators of $\mathbb{Z}_6$ are $1$ and $5$, $\text{Aut}(\mathbb{Z}_6)$ is the set of functions that send $1$ to $1$ or to $5$ and $5$ to $1$ or $5$, or not?
Such a function is the identity function and the function $x\mapsto 5x$, right?
Which function do we use to show the isomorphism?
I think you're in the right track: any homomorphism from any group $\;G\;$ towards any group is completely determined by its action on some generating set of $\;G\;$. If we talk of automorphisms then generators are mapped to generators
In our case, $\;G:=\Bbb Z_6\;$ is a cyclic group of order 6, and thus an automorphism is completely determined by its action on a generator of $\;G\;$.
Since there are two elements of order six here, namely $\;1,5\pmod 6\;$ , every automorphism must map any of them into itself or the other one, and thus clearly Aut$\,(G)\cong\Bbb Z_2\;$