Show that $|\{\varphi\in \text{Aut}(\mathbb{Z}_{15})\mid \varphi=\varphi^{-1}\}|=4$

Question

I want to show that there are exactly $4$ elements of the group $(\text{Aut}(\mathbb{Z}_{15}),\circ)$ with an order such that it divides $2$, that is, an element $\varphi\in\text{Aut}(\mathbb{Z}_{15})$ such that $\varphi=\text{id}_{\mathbb{Z}_{15}}\lor \varphi^{2}=\text{id}_{\mathbb{Z}_{15}}$ which is equivalent to $\varphi=\varphi^{-1}$.

Of course, $\text{id}_{\mathbb{Z}_{15}}$ is one such function, I've been trying to find another one (which must be of order $2$) explicitly but I'm not making much progress. Even if I find the three other functions, I'm not sure how I'm going to prove there are on more than $3$.

I make this post looking for some good hints on how to solve this problem.

Problem

Show that $$\left|\{\varphi\in\text{Aut}(\mathbb{Z}_{15})\mid \varphi=\varphi^{-1}\}\right|=4.$$

Answer 1

Hint: $${\rm Aut}(\Bbb Z_n)\cong U(n),$$

where $U(n)$ is the group of units modulo $n$.

Reference: Gallian's "Contemporary Abstract Algebra (Eighth Edition)", Theorem 6.5.

By definition of $U(15)$, its elements, modulo $15$, are $\{1,2,4,7,-7,-4,-2,-1\}$; then the elements of order at most two are $1,4,-4,-1$.