 Solution

If $|G|=p^{a}m$ 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 $g^{-1}Pg\le Q$; equality follows from their orders. Thus all Sylow $p$-subgroups are conjugate.

Let $P$ now act by conjugation on their set. If it fixes $Q$, then $P\le N_G(Q)$, 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 $\boxed{n_p\equiv 1(\mathrm{mod}p)}$.

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 $H\le Q$. If distinct Sylow subgroups meet trivially, then $H\le P\cap Q=1$. Every other orbit therefore has the full size $|P|$, givingConjugacy also gives $n_p=[G:N_G(P)]$, a divisor of $[G:P]$.