In how many ways can $\mathbb{Z_5}$ act on $\{1,2,3,4,5\}$ In how many ways can $\mathbb{Z_5}$ act on $\{1,2,3,4,5\}$?
I could figure out if $\mathbb{Z_5}$ acts on a set $\{1,2,3,4,5\}$ the orbit $\theta_i$ is either $1$ (the tribial action) or $5$.
Now we see that if the action is transitive then the shouldnt it correspond to $5$ cycles of the form $(12345)^i$ where $1 \le i \le 4$?
So since the homomorphism $\mathbb{Z_5} \to S_5$ is completely determined by $1_5$, it can send the first symbol to any one of $4$ choices the second symbol to any one of the $3$ choices and so on. So there are $4!$ choices.
Theres an answer to the question in the website whoch I was not able to understand. Is this ok?
Also I think we will get the same answer if we replace $\mathbb{Z}_m$ by any other cyclic group, say $\mathbb{Z}$.
 A: A group action $\phi$ is homomorphism from a group $G$ to the symmetric group of a set $X$ i.e. $\phi: G \rightarrow s y m(X)$ is called group action of group $G$ to the set $X$.
We have $G=\mathbb{Z}_{5}$ and $X=\{1,2,3,4,5\}$, Sym $(X)=S_{5}$
So, we have to find number of homomorphisms from $\mathbb{Z}_{5}$ to $S_{5}$.
One is trivial homomorphism in which all elements goes to identity.
For others, $\mathbb{Z}_{5}=\{0,1,2,3,4\} ; O(1)=O(2)=O(3)=O(4)=5$
We have a result if $\phi: G \rightarrow G_{1}$ is a homomorphism, then $O(\phi(x))$ will divide $O(x)$
So $O(\phi(1))|O(1) \Rightarrow O(\phi(1))| 5$
Now, all 5 cycles in $S_{5}$ have order 5 only.
Number of 5 cycles in $S_{5}$ is 24
So $\phi$ (1) has 24 choices.
$\Rightarrow 24$ are the only choices and one is trivial. $\Rightarrow$ Total 25 number of homomorphisms are possible.
A: You can't send $1$ to any element, you can only send it to elements whose oreder divide $5$ as you have already noted, because if $\varphi$ is a homomorphism to $S_5$ then
$$ \varphi(1+1+1+1+1) = \varphi(1)^5 = \varphi(0) = e $$
Now if we want the action to be transitive it is enough for it to be non trivial. So the orbit will be any $5$ cycle, WLOG we can assume the cycle representation to start from $1$, thus $\varphi(1) = (1 a_1 a_2 a_3 a_4)$ now $a_i$ can be arrarnged in any way so there will be $4!$ ways
