Counting Group Actions from a Finite Group to Itself Given a finite group $G$, how many group homomorphisms $G \to \mathrm{Perm}(G)$ are there? Alternatively, in how many ways can a finite group act on itself? 

Here are some food for thoughts, this is far from an answer as I don't know it.


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*If $G$ is cyclic, then $G$ can acts on itself through $|G|!$ different ways. Indeed, if $x\in G$ is a generator of $G$, there are as many homomorphisms from $G$ to $S(G)$ as there are elements in $S(G)$, since a group homomorphism from $G$ is uniquely determined by the image of $x$.

*Let $G_1$, $G_2$ be two finite groups and let $G:=G_1\oplus G_2$. If $\varphi$, respectively $\psi$, is a group homomorphism from $G_1$ to $S(G_1)$, respectively from $G_2$ to $S(G_2)$, then $\varphi\oplus\textrm{id}_{G_2}$, respectively $\textrm{id}_{G_1}\oplus\psi$ is a group homomorphism from $G_1\times G_2$ to $S(G)$. Furthermore, distincts $\varphi$, respectively distincts $\psi$, leads to distincts $\varphi\oplus\textrm{id}_{G_1}$, respectively distincts $\textrm{id}_{G_2}\oplus\psi$. Besides, if $\varphi$ or $\psi$ is not the identity of $G_1$ or $G_2$, then $\varphi\oplus\textrm{id}_{G_2}$ is distinct from $\textrm{id}_{G_2}\oplus\psi$. Therefore, one has: $$|\textrm{Hom}(G,S(G))|\geqslant|\textrm{Hom}(G_1,S(G_1))|\times|\textrm{Hom}(G_2,S(G_2))|-1.$$
I believe this lower bound is far from being optimal.

*If $G$ is finite abelian, then $G$ is a direct product of a finite number of cyclic groups. This follows from the classification of finitely generated modulus on principal ideal domains. If $\displaystyle G=\bigoplus_{i=1}^n\mathbb{Z}/(d_i)$, according to the following point, there is at least $d_1!\cdots d_n!-n+1$ group actions of $G$ on itself. Notice that whenever $G$ is cyclic, this lower bound is reached. It would be an interesting question to ask when this lower bound is reached, but I have no clue at that moment.
