Normal subgroup with index that divides n!

Question

Is this argument valid?

If $G$ is a finite group with $n$ Sylow $p$-subgroups (in particular $n = 1$ mod $p$ and $n$ divides $|G|$), then $G$ permutes them acting by conjugation. Therefore there is a homomorphism $\phi: G \to S_n$ (the symmetric group on $n$ elements). If we consider the kernel of this map, then it is a normal subgroup of $G$ with index $|G|/|\phi(G)|$, which in particular divides $|S_n| = n!$

Answer 1

Hint:

What does the first isomorphism theorem tell you?

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Note that the size of the kernel is $|G|/|\phi(G)|$, not the index of the kernel.