Suppose n is a natural number, at least 3. Is the following true?
Any two subgroups of the alternating group $A_n$ that are conjugate inside the symmetric group $S_n$ (With the natural embedding of $A_n$ in $S_n$) are also conjugate inside $A_n$?
Background: Elements of $A_n$ that are conjugate in $S_n$ need not be conjugate in $A_n$ (there is a criterion for splitting based on the cycle decomposition -- if the cycle type comprises distinct odd cycle sizes, then the conjugacy class splits). However, it turns out that cyclic subgroups of $A_n$ that are conjugate in $S_n$ must be conjugate in $A_n$, because for any two elements of the same cycle type, we can always find a power of one element that is conjugate to the other element (with a little combinatorial manipulation). It's not immediately clear, though, how this can be extended to an arbitrary subgroup.