Number of conjugates of $S_3$ in $S_4$ How to quickly deduct there are 4 conjugates of $S_3$ in $S_4$? Since conjugate subgroups are isomorphic, we can have at least 4 conjugates of $S_3$ in $S_4$. But I'm not sure why there aren't more. 
*I'm aware there are posts that address similar questions but most involve "labeling" schemes or graphs. I'm yet to find a concise approach to this problem.

Let $G = S_4$ act on its subgroups by conjugation.  The conjugates of the standard copy of $S_3$ constitute an orbit for this action.  The size of the orbit is the index of the stabilizer subgroup.  The stabilizer $N$  is what's called the normalizer.  It is the set of $\pi \in S_4$ such that
$\pi S_3 \pi^{-1} = S_3$.  One has $S_3 \subseteq N \subseteq S_4$ just by definition.  Show that this together with the observation that the size of the orbit is at least 4 implies that $S_3 = N$, and the size of the orbit is exactly $4$. 
