$S_6$ contains two subgroups that are isomorphic to $S_5$ but are not conjugate to each other This is a problem from Ph.D. Qualifying Exams. 
Show that the symmetric group $S_6$ contains two subgroups that are isomorphic to $S_5$ but are not conjugate to each other.
Here is my method.  $S_5$ contains 6 Sylow 5-subgroup, and $S_5$ act by  conjugation on the 6 groups transitively by Sylow's 2nd Theorem, therefore induces a homomorphism $\phi: S_5 \to S_6$. Now, if $ker\phi$ is trvial, then $Im\phi$ is a subgroup of $S_6$ isomorphic to $S_5$.  Since $Im\phi$ is a transitive subgroup, it is not conjugate to the subgroup which permutes 5 letters fixing 1 letter, i.e., the subgroup $Sym\{1,2,3,4,5\}\cong S_5$.
So what we need to do is to show $ker\phi$ is trvial. Taken one of 6 Sylow 5-subgroups $H$, there are 6 conjugation orbits, so  $N_G(H)$, the normalizer of $H$ has order $20=\frac{120}{6}$ by counting formula. Here I stopped. I know if I can show the intersection of the 6 normalizers contains only identity element, it will be done, but how can I proceed?
 A: $S_6$ has two conjugacy classes of subgroups isomorphic to $S_5$. First of all the "trivial" one: stabilizers of a point. That gives you 6 subgroups. The second is the conjugacy class of $\langle(1234),(3456)\rangle$. Again $6$ subgroups. The subgroups in these two classes are not conjugate in $S_6$, because the members of the first fix a point, while the members of the second one act transitively. 
A: As requested, here is a summary of the points raised in the comments.
The argument as posted in the question is very nearly complete.  
As is true for any homomorphism, $Ker(\phi)$ is a normal subgroup. Of course, $S_5$ really hasn't got any normal subgroups.  There is $S_5$ itself, the trivial group, and $A_5$.  Now, as was sorted out in the question itself, the fact that all the $5$-Sylow subgroups are conjugate implies that $Im(\phi)$ is non-trivial, it has to have at least $6$ elements!  It follows that $Ker(\phi)≤20$.  That rules out both the complete group $S_5$ and the alternating group $A_5$, so $Ker(\phi)$ must be trivial.  As was pointed out (correctly) in the question this is all that was needed to complete the argument.
