Let $\phi:\Bbb{Z}_{20} \to \Bbb{Z}_{20}$ be an automorphism and $\phi(5)=5$. What are the possibilities of $\phi(x)$?

It's a lot easier to see exactly how flexible the automorphisms are (and count them) if you use $\Bbb Z_{20}\cong \Bbb Z_4\times\Bbb Z_5$, with $5\mapsto (1, 0)$

This is known as the Chinese remainder theorem: If $m, n$ are coprime integers, then $\Bbb Z_{mn}\cong \Bbb Z_m\times \Bbb Z_n$. The most natural isomorphism is $1\mapsto (1, 1)$, so that's the one I'll use here. this gives the $5\mapsto(1, 0)$ above.

An automorphism of $\Bbb Z_4\times \Bbb Z_5$ must send $(1, 0)$ to an element of order $4$ (i.e. $(1, 0)$ or $(3, 0)$) and it must send $(0,1)$ to an element of order $5$ (i.e. $(0,i)$ for $1\leq i\leq 4$), and any automorphism is uniquely determined by where it sends these two elements.

We want all automorphisms which fix $5\in \Bbb Z_{20}$. That corresponds to automorphisms which fix $(1, 0)$ in $\Bbb Z_4\times\Bbb Z_5$. That means the only leeway we have is where $(0,1)$ is sent. We have four options, and they all work. Thus the four maps we are after all map $(1,0)$ to $(1,0)$, and then map $(0,1)$ to either one of the four order-$5$ elements $(0,i)$.

In order to translate back to $\Bbb Z_{20}$ I think it's easiest to see what happens to $(1, 1)$: It is sent to some element $(1, i)$ where $1\leq i\leq 4$. Any such choice gives a valid automorphism. Taking our designated isomorphism back to $\Bbb Z_{20}$, we get the following correspondences between automorphisms: $$ \begin{array}{|c|c|} \hline \text{image of (1, 1)}&\text{image of $1$}\\ \hline (1, 1) & 1\\ (1, 2) & 17\\ (1, 3) & 13\\ (1, 4) & 9\\\hline \end{array} $$

$$5\phi(1) \equiv 5 \pmod{20} \iff \phi(1)\equiv 1 \pmod4 \iff \phi(1)\equiv x \pmod{20} $$ where $x\in \{1, 5, 9, 13, 17\}$. This gives us $5$ automorphisms, each of the form $$\phi(n) = nx $$

But since $5$ is not a generator of $Z_{20}$ hence $\phi(1)\neq 5$ Hence, $\phi(x)=x,9x,13x,17x$ are the only possibilities.