How many group actions are there of $Q_8$ on a 4 element set. The question is to determine how many group actions $\alpha:Q_8\times X→ X$ exist, up to isomorphism, where $|X|=4$. Here is my solution so far:
Call a pair $(X,\alpha)$ a $Q_8$-set if there exists a group action $\alpha:Q_8\times X→ X$.
We know of $4$ actions immediately: The action of $Q_8$ on $\{1,-1,i,-i\},\{1,-1,j,-j\},\{1,-1,k,-k\}$ given by conjugation, and the action of $Q_8$ on $Q_8/\{1,-1\}$ given by left or right translation. The $Q_8$-sets $(Q_8/\{1,-1\},\text{left})$ and $(Q_8/\{1,-1\},\text{right})$ are isomorphic as $Q_8$-sets. The three conjugation actions are distinct (right?). I was trying to show that these are the only four actions.
An action of $Q_8$ on a $4$ element set is equivalent to a homomorphism $Q_8\to S_4$, so it comes down to showing there are $4$ such homomorphisms. No homomorphism can have kernel $\{1\}$, because otherwise $S_4$ has a subgroup isomorphic to $Q_8$, which is not possible. I was tempted to say that,  since there are only $4$ other normal subgroups, there are only $4$ homomorphisms, hence these are all the group actions. But I know that two distinct homomorphisms can have the same kernel — is this still true in this case, or can I argue that (same kernel) implies (same homomorphism) in this specific case? Are there more than $4$ actions? I wanted to do it avoiding the usual way of tracking element orders/etc.. Is this unavoidable?
EDIT: I missed the trivial action — So I'm trying to show that there are five actions, corresponding to all subgroups of $Q_8$ except $\{1\}$.
 A: Your idea to count homomorphisms $Q_8\to S_4$ is a good one, with the caveat that we only want to count them up to conjugacy in $S_4$.
Suppose $h\colon Q_8\to S_4$ is a homomorphism. Then $\ker(h)$ is a normal subgroup of $Q_8$, so it is one of: $\{1\}$, $\{1,-1\}$, $\{1,i,-1,-i\}$, $\{1,j,-1,-j\}$, $\{1,k,-1,-k\}$, or $Q_8$.

*

*If $\ker(h) =  \{1\}$, then $h$ is injective. Since $S_4$ has no subgroup isomorphic to $Q_8$, this case is impossible.


*If $\ker(h) = \{1,-1\}$, then $\mathrm{im}(h)\cong Q_8/\{1,-1\}\cong K_4$. Now there are $4$ subgroups of $S_4$ isomorphic to $K_4$. Three of them are generated by two disjoint transpositions: $\{\varepsilon, (a\,b),(c\,d),(a\,b)(c\,d)\}$. The fourth consists of the identity and all products of two disjoint transpositions: $\{\varepsilon,(1\,2)(3\,4),(1\,3)(2\,4),(1\,4)(2\,3)\}$. Once we've pinned down the image, there are $6$ homomorphisms with that image, corresponding to the $6$ automorphisms of $K_4$ (each non-identity element of $Q/\{1,-1\}$ can be mapped to any element of the image). So in total there are $24$ homomorphisms with kernel $\{1,-1\}$. But we only  want to count actions up to isomorphism. Homomorphisms with image of the form $\{\varepsilon, (a\,b),(c\,d),(a\,b)(c\,d)\}$ correspond two actions in which two of $\pm i,\pm j,\pm k$ act as disjoint transpositions and the third acts as the product of the two disjoint transpositions. Up to isomorphism there are $3$ of these, determined by which of $\pm i,\pm j,\pm k$ acts as a product of two disjoint tranpositions. Homomorphisms with image $\{\varepsilon,(1\,2)(3\,4),(1\,3)(2\,4),(1\,4)(2\,3)\}$ correspond to just one action up to isomorphism, in which each non-identity element acts as a product of disjoint transpositions. So there are $4$ actions with kernel $\{1,-1\}$, up to isomorphism.


*If $\ker(h) = \{1,i,-1,-i\}$, then $\mathrm{im}(h)\cong Q_8/\{1,i,-1,-1\}\cong C_2$. Now a homomorphism $C_2\to S_4$ is determined by an element of $S_4$ of order $2$. There are $9$ such elements: $6$ transpositions $(a\,b)$, and $3$ elements which are products of two disjoint transpositions $(a\,b)(c\,d)$. In each case, all homomorphisms give isomorphic actions, but the two cases correspond to  non-isomorphic actions. So we have $9$ homomorphisms and $2$ actions up to isomorphism in this case.


*The same analysis works if $\ker(h) = \{1,j,-1,-j\}$ or $\{1,k,-1,-k\}$. So there are $9$ more homomorphisms and $2$ more actions up to isomorphism in each of these cases.


*If $\ker(h) = Q_8$, $h$ is the trivial homomorphism. This corresponds to the trivial action.
In total, there are $24+9+9+9+1 = 52$ homomorphisms $Q_8\to S_4$ and $4+2+2+2+1 = 11$ actions up to isomorphism.
