Two Sylow $p$-subgroups with the same image are conjugate by an element of the kernel

I need help with the following exercise, which, judging by its position in the book, should follow more or less directly from the Sylow theorems.

Let $G,G'$ be two finite groups and $\phi$ a homomorphism of $G$ onto $G'$. Let $p$ be a prime number and $P,P_1$ two Sylow $p$-subgroups of $G$ such that $\phi(P)=\phi(P_1)$. Show that there exists $x\in\text{Ker}(\phi)$ such that $P=xP_1x^{-1}$.

Well, I know that there exists $y\in G$ such that $P=yP_1y^{-1}$ and it's easy to see that the set of $x$ such that $P=xP_1x^{-1}$ is $yN_G(P_1)$. So we have to show $$yN_G(P_1)\cap\text{Ker}(\phi)\neq\emptyset.$$

But what now? A hint would be most welcome.

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Even though it is early in the book, it is still a new, hard, and important idea: use the Sylow theorems in subgroups containing both Sylow subgroups. Jyrki's answer is exactly it, and a very important technique. – Jack Schmidt Oct 19 '11 at 20:50

Hint: Recall that if $N\unlhd G$ and $K\le G$, then the set of products $NK=\{nk\mid n\in N, k\in K\}$ is also a subgroup of $G$. The condition $\phi(P)=\phi(P')$ means that the subgroups $(\ker\phi) P$ and $(\ker\phi) P'$ are equal. Let $H$ be that subgroup of $G$. What do the Sylow theorems tell you about the Sylow $p$-subgroups of $H$? Can you list a few of those?