Find group with distinct Sylow $p$-subgroups that share a normalizer

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...

• 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 $H_1\unlhd N_G(H_1)$ :-) – Jyrki Lahtonen Dec 11 '17 at 20:06
• A bit more seriously. This argument is often seen when proving that $N_G(N_G(P))=N_G(P)$ for all Sylow subgroups $P$ of $G$. – Jyrki Lahtonen Dec 11 '17 at 20:07
• @JyrkiLahtonen Please move your comment to an answer so that this question does not show as unanswered. Thanks. – Stephen Meskin Dec 11 '17 at 20:54
• @StephenMeskin James already gave that as an answer (obviously he arrived at it independently from me), but then deleted his answer. I could use my powervote undelete, but I'm a bit hesitant to do that against his wish. – Jyrki Lahtonen Dec 11 '17 at 21:28
• @JyrkiLahtonen If you won't post the answer, I will as a community wiki. – Stephen Meskin Dec 11 '17 at 21:34

"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 $H_1⊴N_G(H_1)$."