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$mathbb Z_{5}$ is the group to act on the set ${ 1,2,3,4,5}$. In how many ways is that possible $?$
Now $0$ will give the identity map. $1$ will give a bijection in $5!$ ways so will the others and the number of possible bijection being $5!$ the bijections given by $1$ will coincide with those given by others. All gets mixed up here.

An equivalent way of thinking of group actions is as follows: a group $G$ acting on a (finite) set $X$ is exactly a homomorphism $$theta:Gto S_{|X|}$$
where $S_{|X|}$ is the symmetric group on $|X|$ elements, and $theta(g)$ is the permutation of $X$ given by $xmapsto g(x)$. And every such homomorphism gives a group action: define $g(x)$ to be where $theta(g)$ sends $x$.

How many homomorphisms are there from $mathbb Z/5mathbb Zto S_5$?

$mathbb Z/5mathbb Z$ is cyclic, so any homomorphism will be completely determined by where it sends the generator $1$. But $1$ has order $5$ in $mathbb Z/5mathbb Z$, so it can be sent only to elements with orders dividing $5$.

So we just need to find the number of elements of orders $1$ and $5$ in $S_5$. Can you finish from here?

In this answer, let $C_m$ denote the cyclic group of order $m$ for $minmathbb{Z}_{>0}$. I also write $C_0$ for the infinite cyclic group.

There is one obvious group action: the trivial action. Now, we consider a nontrivial action. Then, there exists an element of ${1,2,3,4,5}$, say $1$, not fixed by $C_5$. However, the size of the orbit of $1$ would then be $5$ (since the size of the orbit divides the order of $C_5$ and it is not $1$). Hence, this action is transitive as the orbit must be the whole ${1,2,3,4,5}$. Therefore, this action is a permutation of ${1,2,3,4,5}$ on a circle (where $1in C_5$ is rotation by $frac{2pi}{5}$). On the other hand, every permutation of ${1,2,3,4,5}$ on a circle can be made an action of $C_5$ on this set. There are $(5-1)!=24$ such permutations. Hence, in total, there are $1+24=25$ possible actions of $C_5$ on ${1,2,3,4,5}$.

In how many ways can the group $C_6$ act on ${1,2,3,4,5,6}$?

Hint: The answer is $1+15+40+45+120+15+40+120=396$.

In general, if $a^m_n$ is the number of ways the group $C_m$ can act on the set ${1,2,ldots,n}$, where $m$ and $n$ are nonnegative integers, then
$$sum_{n=0}^infty,a^m_n,frac{t^n}{n!}=expleft(sum_{substack{{dinmathbb{Z}_{>0}}\{dmid m}}},frac{t^d}{d}right),.$$ In particular, it can be shown that $a^0_n=n!$ for every $ninmathbb{Z}_{geq 0}$, or
$$sum_{n=0}^infty,t^n=sum_{n=0}^infty,a^0_n,frac{t^n}{n!}=expleft(sum_{substack{{dinmathbb{Z}_{>0}}\{dmid 0}}},frac{t^d}{d}right)=expleft(sum_{d=1}^infty,frac{t^d}{d}right)=expBigg(lnleft(frac{1}{1-t}right)Bigg)=frac{1}{1-t},.$$