= Solution
For a finite <group> with $|G|=p^am$ and $p\nmid m$, a <Sylow p-subgroup> has order $p^a$. The <Sylow theorems> assert existence, containment of every <p-subgroup> in a Sylow subgroup, conjugacy of all Sylow subgroups, and $n_p\mid m$, $n_p\equiv1\pmod p$.
For <Sylow subgroups of A6>, $|A_6|=6!/2=360=2^3\cdot3^2\cdot5$. Explicit choices are
$$
\boxed{P_5=\langle(12345)\rangle\cong C_5,\qquad P_3=\langle(123),(456)\rangle\cong C_3\times C_3.}
$$
Their elements are even permutations, and their orders are the largest powers of five and three dividing $360$.
Every nonidentity element of $P_5$ is a five-cycle. There are $\binom65\,4!=144$ five-cycles, and distinct subgroups of order five meet only at the identity. Each contributes four such elements, giving
$$
\boxed{n_5=144/4=36.}
$$
For $P_3$, each unordered partition of the six letters into two triples gives the subgroup generated by the two triple-cycles. It is independent of the chosen orientation of either cycle. There are $\tfrac12\binom63=10$ such partitions. Different partitions give different subgroups, since the four single three-cycles in a subgroup identify the two triples. Every Sylow subgroup of order nine is conjugate to the displayed $P_3$, so every one arises from a partition in this way. Hence
$$
\boxed{n_3=10.}
$$
The <normalizer> is $N_G(H)=\{g\in G:gHg^{-1}=H\}$. Its definition makes $H$ normal in $N_G(H)$. If $H$ is a <Sylow subgroup> of $G$, it is also a Sylow subgroup of $N_G(H)$, since $H\leq N_G(H)\leq G$ leaves no room for a larger power of $p$. The <Sylow theorems> make all Sylow subgroups of the normalizer conjugate there, and normality of $H$ makes every such conjugate equal to $H$. Thus \b[$H$ is the unique Sylow subgroup of its normalizer].
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