show/hide this revision's text 3 added 2 characters in body

Not sure if this question really qualifies for MO.

Anyway, the answer very much depends on the group $G$. In most cases $n_p(G)\ne n_q(G)$ for distinct prime divisors of the group order. However, there are infinitely many examples very where equality occurs: If $r$ is an odd prime, then $n_p(\text{PSL}(2,r))=r(r+1)/2$ for each odd prime divisor $p$ of $r-1$.

But there are other examples too. For instance the atlas of finite simple groups shows that in the Janko group $J_1$, the normalizer normalizers of the $3$-Sylows and $5$-Sylows have order $60$.
show/hide this revision's text 2 added 5 characters in body

Not sure if this question really qualifies for MO.

Anyway, the answer very much depends on the group $G$. In most cases $n_p(G)\ne n_q(G)$ for distinct prime divisors of the group order. However, there are infinitely many examples very equality occurs: If $r$ is an odd prime, then $n_p(\text{PSL}(2,r))=r-1$ n_p(\text{PSL}(2,r))=r(r+1)/2$ for each odd prime divisor $p$ of $r-1$.

But there are other examples too. For instance the atlas of finite simple groups shows that in the Janko group $J_1$, the normalizer of the $3$-Sylows and $5$-Sylows have order $60$.
show/hide this revision's text 1

Not sure if this question really qualifies for MO.

Anyway, the answer very much depends on the group $G$. In most cases $n_p(G)\ne n_q(G)$ for distinct prime divisors of the group order. However, there are infinitely many examples very equality occurs: If $r$ is an odd prime, then $n_p(\text{PSL}(2,r))=r-1$ for each odd prime divisor $p$ of $r-1$.

But there are other examples too. For instance the atlas of finite simple groups shows that in the Janko group $J_1$, the normalizer of the $3$-Sylows and $5$-Sylows have order $60$.