I want to show that $\text{Aut}(\mathbb{Z}_8)$ is isomorphic to $\mathbb{Z}_2\times\mathbb{Z}_2$.
The group $\mathbb{Z}_8$ is cylcic and is generated by one element. The possible generators are $1,3,5,7$.
Each automorphism maps each of the element of the set $\{1,3,5,7\}$ to one of the element $\{1,3,5,7\}$, right?
But how can we show that that the automorphism group of $\mathbb{Z}_8$ is isomorphic to $\mathbb{Z}_2\times\mathbb{Z}_2$ ?
An automorphism $f:\Bbb Z_8 \to \Bbb Z_8$ is determined by $f(1)$ (prove it). Since $1$ is a generator and generators must map to generators, $f(1)\in \{1,3,5,7\}$. So there are 4 automorphisms. Since the only groups of order 4 are $\Bbb Z_4$ and $\Bbb Z_2 \times \Bbb Z_2$, you just have to show that $Aut(\Bbb Z_8)$ has no element of order 4.