Let $X$ be a finite set. Determine the number of actions of $\mathbb Z$ on $X$.
If $X$ is a finite set with $|X|=m$, then $|\{f:X \to X : \text{f is bijective}\}|=m!$.
Finding the number of actions is counting the number of morphisms $\phi:\mathbb Z \to Sym(X)$. By the condition $\phi(1)=Id_{Sym(x)}$ and the fact that $\phi$ is a morphism, we have $\phi(n)=\phi(1+...+1)=Id_{Sym(X)} \circ ... \circ Id_{Sym(X)}=Id_{Sym(X)}$, so there is an only possible action which is the "trivial" action, which is $g.x=x$ for all $g \in \mathbb Z$, $x \in X$.
I would like to verify if my solution is correct.
As said by @anon, your solution is incorrect. In some sense you had a fruitful idea by focussing on $\phi(1)$, but you applied that focus incorrectly by confusing additive and multiplicative notation.
Instead, use the fact that for any group $G$ with group operation denoted $g \cdot h$, a homomorphism $\phi : \mathbb{Z} \to G$ is completely determined by the value of the single element $\phi(1) \in G$, because $$\phi(n) = \phi(\,\underbrace{1 + \cdots + 1}_{\text{($n$-times)}}\,) = \underbrace{\phi(1) \cdot \ldots \cdot \phi(1)}_{\text{($n$-times)}} = \phi(1)^n $$
Can you take it from there?