There exist Sylow subgroups $P$ and $Q$ for which $[P:P \cap Q]=[Q : P \cap Q] = p$. From D&F's sylow theory section:

Show that if $n_p\not\equiv 1 \mod p^2$ then there are distinct Sylow $p$-subgroups $P$ and $Q$ of $G$ for which $[P:P \cap Q]=[Q : P \cap Q] = p$.

Are automorphism groups required for this question?  Does it involve permutation represntations or does it follow directly from the Sylow theorems?

Consider the action (by conjugation) of $P$ on the set $A$ of all Sylow $p$-subgroups of $G$. Its orbits have the length $p^k, k\ge 0$. Since $|A|=n_p\not\equiv 1 \mod p^2$ and there is just one single-element orbit (namely, $P$), there exists an orbit $B$ of length $p$. If $Q\in B$, the stabilizer of $Q$ is a subgroup $H\subset P$ with $[P:H]=p$. Since $h^{-1}Qh=Q$ for $h\in H$, then $H$ lies in the normalizer of $Q$, hence $H\subset Q$ and $H=P\cap Q$.
