A homomorphism from a group $G$ of order $6$ to $S_4$ is equivalent to an action of $G$ on the set $X=\{1,2,3,4\}$. By the orbit-stabilizer theorem, every orbit must have size either $1$, or $2$, or $3$ (because $4\nmid 6$). Therefore, the only allowed orbit equations are:
- $\space\space\space 4 = 1+1+1+1$
- $\space\space\space4 = 2+1+1$
- $\space\space\space4 = 2+2$
- $\space\space\space4 = 3+1$
Am I right if I consider as upper bound of the number in the title the number of ways the orbit equations 1 to 4 can actually be realized? I mean:
1 can be realized in the only way: $$\{\{1\},\{2\},\{3\},\{4\}\}$$ 2 in the following $6$ ways: \begin{alignat}{1} &\{\{1,2\},\{3\},\{4\}\} \\ &\{\{1,3\},\{2\},\{4\}\} \\ &\{\{1,4\},\{2\},\{3\}\} \\ &\{\{2,3\},\{1\},\{4\}\} \\ &\{\{2,4\},\{1\},\{3\}\} \\ &\{\{3,4\},\{1\},\{2\}\} \\ \end{alignat} 3 in the following $6$ ways: \begin{alignat}{1} &\{\{1,2\},\{3,4\}\} \\ &\{\{1,3\},\{2,4\}\} \\ &\{\{1,4\},\{2,3\}\} \\ &\{\{2,3\},\{1,4\}\} \\ &\{\{2,4\},\{1,3\}\} \\ &\{\{3,4\},\{1,2\}\} \\ \end{alignat} and 4 in the following $4$ ways: \begin{alignat}{1} &\{\{2,3,4\},\{1\}\} \\ &\{\{1,3,4\},\{2\}\} \\ &\{\{1,2,4\},\{3\}\} \\ &\{\{1,2,3\},\{4\}\} \\ \end{alignat} So, I'd say that there are at most $1+6+6+4=17$ homomorphisms as in the title (the trivial one included), but according to this answer (where $G=C_2\times C_3$) this conclusion is wrong. Does it mean that distinct actions can have one same orbit setup?
Indeed distinct actions can have the same orbits. For instance, on any group left multiplication $a_g(h)=gh$ and inverse right multiplication $b_g(h)=hg^{-1}$ are two potentially different actions, and they both have the entire group as one orbit.