Prove *by group actions* that there doesn't exist any normal subgroup $H$ such that $S_5/H $ is isomorphic to $S_4$ I was trying to give an answer to this very same question by means of group actions (so I think this is not a duplicate, at least as long as the answers therein do not use group actions).
My attempt: If such a $H$ exists, then there is a homomorphism from $S_5$ onto $S_4$ with kernel $H$. But then, there is an action of $S_5$ on $X:=\{1,2,3,4\}$ with kernel $H=\bigcap_{i=1}^4\operatorname{Stab(i)}$. $X$'s partitioning into orbits can be any of the following:
a) $4=1+1+1+1$,
b) $4=1+1+2$,
c) $4=2+2$, and
d) $4=4$.
By the Orbit-Stabilizer Theorem, case a) corresponds to $\operatorname{Stab(i)}=S_5, \forall i\in X$, whence $H=S_5$: contradiction; cases b) and c) correspond to $|H|=5!/2=60$: contradiction. Finally, case d) corresponds to a transitive action and hence the $4$ stabilizers are conjugate in $S_5$.

How can I conclude from here?

Edit. I've just realized that also the case $4=1+3$ must be addressed.
Edit#2. Also the cases b) and c) are not so plain as I thougth, since the subgroups of index $2$ might not be unique, in principle.
 A: Let me have a punt at this. Sounds like you are looking for a proof with a different flavour than the other similar post. Perhaps this will do it.
You've set it up nicely with an onto homomorphism $\phi:S_5\rightarrow S_4$ with kernel $H$. This subgroup $H$ has order 5!/4! = 5, hence it is a cyclic subgroup. Now we play hunt of the contradiction. The fact that every non-trivial member of $H$ has to have order 5 is strong and will give us something to try and contradict with.
We know $S_4$ acts transitively on the set of four elements, so if this homomorphism exists, then $S_5$ acts transitively on the set of four elements too using the induced action of $\phi$. By this I mean the action of $g\in S_5$ on {1,2,3,4} is taken to be the action of $\phi(g)$. Keep in mind this action might be non-standard, eg the cycle (123)$\in S_5$, say, might map to some totally different cycle in $S_4$.
Now $S_4$ has transpositions, so take any $g\in S_5$ mapping to one of these transpositions. So $\phi(g)$ has order 2, so $g^2$ lies in the kernel $H$. So, because $H$ was cyclic of order 5, $g^2$ has order 5 or $g^2$ is the identity. If $g^2$ has order 5 then $g$ has order 10, which is a contradiction as no element of $S_5$ has order 10. Hence $g^2$ is the identity, ie $g$ is also a transposition.
Next bit, I'd like to show that $\phi$ is injective on the transpositions. Let $g$ and $h$ be transpositions in $S_5$ with $\phi(g) = \phi(h)$. Then $gh^{-1}\in H$. Again either $gh^{-1}$ has order 5 (which it doesn't) or it is the identity, which implies $g=h$. Hence injective on the transpositions.
Now the contradiction: there are fewer transpositions in S_4 than S_5, and we've just shown $\phi$ is an injective map from the larger to the smaller set of transpositions.
Let me know if I've made a mistake or anything not clear.
