Here a question I thought of, but can't find an answer to
Find two distinct Sylow $p$-subgroups (of a given $p$) $H_1$ and $H_2$ of $G$ such that $N_G(H_1) = N_G(H_2)$.
I don't know if it's actually possible, so I should qualify the question with
If no such pair exists, show why.
Well, the easiest case would be if $H_1$ and $H_2$ were normal, however this would imply that $n_p = 1$. Hence, we'd only have one Sylow $p$-subgroup.
My intuition says that such a pair does exist, however. Of course, it's merely intuition...