Find elements of a group $\operatorname{Aut}(\mathbb{Z_{20}})$ automorphisms of cyclic group $\mathbb{Z_{20}}$ 
Find elements of a group $\operatorname{Aut}(\mathbb{Z_{20}})$ automorphisms of cyclic group $\mathbb{Z_{20}}$. Is $\operatorname{Aut}(\mathbb Z_{20})$ cyclic group?

$\mathbb{Z_{20}} = \left\{0,1,...,19\right\}$
We know that an automorphism is an isomorphism from group G into the same group.
Let $\phi$ be random element of $\operatorname{Aut}(\mathbb{Z_{20}})$. Group $\mathbb{Z_{20}}$ is generated by $1$ and $\phi$ is a group homomorphism so $\phi$ is uniquely designated via image on element $1$.
Automorphism $\phi$ doesn't change group order so $\phi(1)$ must be of order $20$. So $|\operatorname{Aut}(\mathbb{Z_{20}})| = 20$.
These are my only observations but I don't know what to do next to get to specific elements
 A: $\operatorname{Aut}(\Bbb Z_{20})\cong\Bbb Z_{20}^×\cong(\Bbb Z_5×\Bbb Z_4)^×\cong\Bbb Z_5^××\Bbb Z_4^×\cong\Bbb Z_4×\Bbb Z_2$ and thus is not cyclic.
A: Hint:
An automorphism of $\mathbf Z/20\mathbf Z$ maps the  generator $1$ on to another generator, so you have to determine the other generators.
Now for an element $g$ with order $r$ in a commutative group $G$ (in multiplicative notation), $g^k$ has order $\frac r{\gcd(r,k)}$, so for $g^k$  to be another generator, we need to have $k$ and $r$ coprime. In the present case the means that $k$ has to be not divisible by $2$ nor $5$.
There are in all $8$ elements $<20$ coprime to $20$., so the group of automorphism you're after has order $8$. Up to isomorphism, there are exactly $5$ groups of order $8$: the cyclic group $\mathbf Z/8\mathbf Z$, the abelian groups $\mathbf Z/2\mathbf Z\times\mathbf Z/4\mathbf Z$ and $\mathbf Z/2\mathbf Z\times\mathbf Z2\mathbf Z$ and two non-commutative  groups: the dihedral group $D_4$ and the quaternionic group $Q_8$. 
Can you say which group $\operatorname{Aut}(\mathbf Z/20\mathbf Z)$ is?
A: Multiplication by $-1$ and by $9$ induce automorphisms of $\mathbb Z / 20 \mathbb Z$. Because
$$(-1)^2 = 1, \quad 9^2 \equiv 1 \pmod{20} \,,$$
both are involutions. The group $\operatorname{Aut}(\mathbb Z / 20 \mathbb Z)$ contains at least two elements of order $2$, so it cannot be cyclic.
