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 $60$.

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$60$. 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\$.