How many different ways can $\mathbb{Z}_3$ act on the set $\{1, 2, 3, 4\}$
This is my attempted proof.
Proof: Any action of $\mathbb{Z}_3$ on the set $\{1, 2, 3, 4\}$ is equivalent to a homomorphism $\phi : \mathbb{Z}_3 \to S_4$. So the question is equivalent to asking how many homomorphisms there are from $\mathbb{Z}_3$ to $S_4$.
Note that since $\mathbb{Z}_3$ is cyclic, any homomorphism $\phi : \mathbb{Z}_3 \to S_4$ is completely defined by $\phi(1)$. But also recall that the order of $\phi(1)$ must divide the order of $1$ in $\mathbb{Z}_3$ which is $3$. Thus $o(\phi(1))$ is either $1$ or $3$. The only element of order $1$ in $S_4$ is $1_{S_4}$ and if $\phi(1) = 1_{S_4}$ then $\phi$ is the trivial homomorphism.
Now observe that there are $$\frac{4!}{(4-3)!3} = \frac{24}{3} = 8$$ elements of order $3$ in $S_4$ which corresponds to $8$ possible non-trivial homomorphisms from $\mathbb{Z_3}$ to $S_4$. Thus there are $1 + 8 = 9$ possible homomorphisms from $\mathbb{Z}_3$ to $S_4$ resulting in $9$ possible actions of $\mathbb{Z}_3$ on $\{1, 2, 3, 4\}$. $\square$
Is my proof correct?
Edit: One of the hints that was given was to consider orbit decompositions and apply the Orbit-Stabilizer Theorem, though I'm not sure how one can use that to prove the above. Could someone outline a proof using this method if possible?
Here is another way at it, just for kicks:
Note that your homomorphism maps $0\mapsto1$, $1\mapsto\psi_1$, and $2\mapsto\psi_2$ such that $\psi_1^2=\psi_2$ and $\psi_2^2=\psi_1$ due to the group action axioms, i.e., $\psi_1^4=\psi_2^2=\psi_1$ so that $\psi_1=1$ or has order $3$. Note that $\psi_2$ is determined by $\psi_1$. There are $8$ elements in $S_4$ of order $3$. Therefore, the number of homomorphisms is $1+8=9$.