Group acting on its set of subgroups by conjugation 
I'm pretty sure for the first $H$, the Stabiliser is all of $S_4$ due to the normality of $V_4$, and so the Orbit is just $V_4$.
For the second $H$, I have that the Stabiliser is $H$, as $4$ has to be left invariant. By Orbit-Stabiliser theorem, the Orbit must be a set of 4 subgroups, which I understood to be the 4 different symmetric groups of 3 elements within $S_4$.
However $H = \langle(1234)\rangle$ is giving me a real headache. I can't seem to get the Stabiliser as easily this time around, as nothing appears to work besides the identity.
 A: Conjugacy classes in the symmetric group correspond to cycle-types. Thus, for $H = \langle (1234)\rangle$, the conjugacy class of $(1234)$ is just $\{(1234),(1243),(1324),(1342),(1423),(1432)\}$. We see that the cycles $(1234),(1432)$ generate the same group. Similarly, so do the cycles $(1243),(1342)$, and also the cycles $(1324),(1423)$.
Thus, the orbit of $H = \langle (1234)\rangle$ are the three groups $H, \langle (1243)\rangle,\langle (1324)\rangle$.
By orbit-stabilizer, the stabilizer $S$ is thus a subgroup of index 3, hence order 8. Clearly $S$ contains $H$, so to describe $S$, it suffices to find an element of $S$ not in $H$.
Note that every element of $H$ commutes with itself. On the other hand, we know that $(1234)$ and $(1432)$ are conjugate, so they must be conjugate by an element of $S_4$ not in $H$. Finding this conjugating element comes down to realizing that given two permutations $\sigma,\tau$, if $\tau(a) = b$, then $(\sigma\tau\sigma^{-1})(\sigma(a)) = \sigma(b)$. Thus, the cycle notation of $\sigma\tau\sigma^{-1}$ is precisely just $\sigma$ applied to the cycle notation of $\tau$. From this we see that the transposition $(24)$ has the property that $(24)(1234)(42) = (1432)$. Thus, the stabilizer $S$ is the group generated by $(1234)$ and $(24)$.
