= Solution
If $|G|=p^am$ with $p\nmid m$, a <Sylow subgroup> for $p$ has order $p^a$. Let $P,Q$ be two such <subgroups>. The action of $P$ by left translation on $G/Q$ has <orbits> of powers of $p$, while $|G/Q|=m$ is prime to $p$. There is a fixed coset $gQ$, giving $g^{-1}Pg\leq Q$; equality follows from their orders. Thus \b[all Sylow $p$-<subgroups> are conjugate].
Let $P$ now act by conjugation on their set. If it fixes $Q$, then $P\leq N_G(Q)$, so $PQ$ is a <subgroup> of $p$-power order and must equal $Q$; hence $P=Q$. The unique fixed point is $P$ itself and all other <orbits> have size divisible by $p$. Consequently $\boxed{n_p\equiv1\pmod p}$.
For the strengthened conclusion, the <stabilizer> of $Q\ne P$ is $H=P\cap N_G(Q)$. The product $HQ$ is a <subgroup> of $p$-power order, so $H\leq Q$. If distinct <Sylow subgroups> meet trivially, then $H\leq P\cap Q=1$. Every other <orbit> therefore has the full size $|P|$, giving
$$
\boxed{n_p\equiv1\pmod{|P|}.}
$$
Conjugacy also gives $n_p=[G:N_G(P)]$, a divisor of $[G:P]$.
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