This is a question from the book Contemporary Abstract Algebra, specifically chapter 6 on isomorphisms. The question is:

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

**Attempt of Solution**

I have made an attempt at solving it. Firstly, we know that an isomorphism carries the identity of the first group to the identity of the second group. Since the identity is $0$, and it is an automorphism, so $\phi(0) = 0$.

Next up, we know that an isomorphism preserves the order of an element. So I made a list of all elements of $Z\_{20}$ with the same order. For reference, here's the list (excluding $5$ and $0$):
\begin{align\*}
1) & \, \{1, 3, 7, 9, 11, 13, 17, 19 \} &\text{ (Order is 20) } \\
2) & \, \{2, 6, 14, 18 \} &\text{ (Order is 10) } \\
3) & \, \{4, 8, 12, 16 \} &\text{ (Order is 5) } \\
4) & \, \{10 \} &\text{ (Order is 2) } \\
5) & \, \{15 \} &\text{ (Order is 4) } \\
\end{align\*}
So we can restrict $\phi$ in that if it is an automorphism, it must map $0, 5, 10$ and $15$ to themselves. Also, it can only map the rest of the elements to some element in their set otherwise order will change.

That give $8! \cdot 4! \cdot 4!$ possibilities in total. Is this approach correct, or I'm missing something else?