TITLE: Find group with distinct Sylow $p$-subgroups that share a normalizer
QUESTION [5 upvotes]: 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...

REPLY [2 votes]: According to a comment by @JyrkiLahtonen and others:
"This is not possible. The normalizer of a Sylow $p$- subgroup has a unique Sylow $p$-subgroup. Well, because the one it has is normal, that is $H_1⊴N_G(H_1)$."