Finding the kernel of an action on conjugate subgroups I'm trying to solve the following problem:

Let $G$ be a group of order 12. Assume the 3-Sylow subgroups of $G$ are not normal.  Prove that $G\cong A_4$.

Here's my attempt: let $\mathscr S$ be the set of 3-Sylow subgroups of $G$. Since the elements of $\mathscr S$ are not normal, by Sylow's theorem, $\# \mathscr S > 1$. Again by Sylow's theorem, $\#\mathscr S = 4$ and the elements of $\mathscr S$ are conjugate to each other.  Hence, one can define a group action of $G$ on $\mathscr S$ by conjugation, and this defines a homomorphism $\phi : G\rightarrow \mathrm{Sym}(\mathscr S)\cong S_4$.  Thus, it suffices to show that $\phi$ is injective and its image is $A_4$.
But I'm stuck at this last step.  I tried to find the kernel of $\phi$ and found that $a\in\mathrm{Ker}\phi\Leftrightarrow \forall H\in\mathscr S\ aHa^{-1} = H$, but I do not understand what this leads to.
I would be most grateful if you could provide a clue (not necessarily a complete solution).

If you are still asking why the image of $G$ is (contained in) $A_4$, consider the homomorphism $G \to S_4 \to \mathbb{Z}/2$, where the latter map is the parity morphism. If it were surjective, there would be a normal subgroup of order 6 in $G$. This leads to a contradiction.
