Let $H$ be a normal subgroup of a finite group $G$ and $p$ a prime number, show that the orbits of the conjugation action induced by $H$ on $X = \{P \leq G \mid P \in Syl_p(G) \}$ all have the same size.
I tried finding a bijection between the orbits $HP$ and $HP'$ for $P,P' \in X$:
$$\phi:HP \to HP' \\ \quad hPh^{-1} \rightarrow hP'h^{-1}$$
But I can't show that $\phi$ is one-to-one; I tried to use that $P$,$P'$ are conjugates but that didn't seem helpful. Any suggestions?
EDIT: Let $P,Q$ be $p$-Sylow subgroups of $G$ and $x \in Stab(P); y \in Stab(Q)$ $\quad$($P=aQa^{-1}$)
$$\begin{align} xPx^{-1}=P &\iff xaQ(xa)^{-1}=aQa^{-1} \\ &\iff a^{-1}xaQ(a^{-1}xa)^{-1}=Q \\ &\iff a^{-1}Stab(P) a \subseteq Stab(Q) \\ yQy^{-1}=Q &\iff ya^{-1}Pay^{-1}=a^{-1}Pa \\ &\iff aya^{-1}P(aya^{-1})^{-1}=P \\ &\iff aStab(Q)a^{-1} \subseteq Stab(P) \\ &\iff Stab(Q) \subseteq a^{-1}Stab(P)a \end{align}$$
But now as $|HP|=(H:Stab(P));|HQ|=(H:Stab(Q))$ and $Stab(Q), Stab(P)$ have the same cardinality because they are conjugates, its orbits should have the same size. Is this enough? Shouldn't the normality of $H$ be used?
You implicitly use the normality of $H$ in your calculations, because in general you do not have that if $gAg^{-1}=B$, then $gN_K(A)g^{-1}=N_K(B)$. You only get that if $gKg^{-1}=K$. What we have is:
Lemma. Let $G$ be a group, let $K_1$ and $K_2$ be subgroups, and let $g\in G$. Then $$gN_{K_1}(K_2)g^{-1} = N_{gK_1g^{-1}}(gK_2g^{-1}).$$
Proof. Let $a\in N_{K_1}(K_2)$. If $gxg^{-1}\in gK_2g^{-1}$, $$ (gag^{-1})(gxg^{-1})(gag^{-1})^{-1} = g(axa^{-1})g^{-1}\in gK_2g^{-1},$$ because $axa^{-1}\in K_2$. In addition, $a\in gK_1g^{-1}$. Therefore, $$gN_{K_1}(K_2)g^{-1}\subseteq N_{gK_1g^{-1}}(gK_2g^{-1})\text{ for all }g\in G.$$ To prove the other inclusion we use this inclusion applied to the normalizer of $gK_2g^{-1}$ in $gK_1g^{-1}$ $$g^{-1}N_{gK_1g^{-1}}(gK_2g^{-1})g \subseteq N_{g^{-1}gK_1g}(g^{-1}gK_2g^{-1}g) = N_{K_1}(K_2).$$ Left multiplying by $g$ and right multiplying by $g^{-1}$ we obtain $$N_{gK_1g^{-1}}(gK_2g^{-1}) \subseteq gN_{K_1}(K_2)g^{-1},$$ proving the equality.
Now let $P$ and $Q$ be two Sylow $p$-subgroups of $G$. Let $g\in G$ be such that $gPg^{-1}=Q$. Then $$N_H(Q) = N_H(gPg^{-1}) = N_{gHg^{-1}}(gPg^{-1}) = gN_H(P)g^{-1},$$ where the first equality holds beause $gPg^{-1}=Q$, the second because $H\triangleleft G$, and the third by the Lemma.
Note also that $gN_H(P)g^{-1}\leq gHg^{-1}=H$. That means that the automorphism of $H$ induced by conjugation by $g$ maps $N_H(P)$ to $N_H(Q)$, and hence the index of $N_H(P)$ in $H$ equals the index of $N_H(Q)$ in $H$. (This equality holds in cardinality even for infinite subgroups, though of course the use of Sylow Theory means we are workign with finite groups here...)
Applying the Orbit-Stabilizer Theorem, we have: $$ \#(\text{orbit of }P) = [H:N_H(P)] = [H:N_H(Q)] = \#(\text{orbit of }Q),$$ proving the equality.