let $G$ be a finite nonabelian simple group.If there exist $p$ and $q$ which are different prime numbers of $G$ such that $n_p(G)=n_q(G)$?
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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 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 $r1$. But there are other examples too. For instance the atlas of finite simple groups shows that in the Janko group $J_1$, the normalizers of the $3$Sylows and $5$Sylows have order $60$. 

