Let $G$ and $G'$ be two finite simple groups and $p$ be a prime divisor of $\vert G\vert$ and $\vert G'\vert$. Also suppose that every Sylow psubgroup of $G$ and $G'$ is a prime order subgroup($C_{p}$. If the number of Sylow psubgroups of $G$ is equal to the number of Sylow psubgroups of $G'$, then can we say that $G\cong G'$?
1 Answer
A good place to look for counterexamples might be nonisomorphic simple groups of the same order.
And indeed we find that $A_8$ and ${\rm PSL}(3,4)$ both have $960$ Sylow $7$subgroups.
The next such pair is ${\rm PSp}(6,3)$ and ${\rm P}\Omega(7,3)$ and they both have the same numbers of Sylow $p$subgroups for $p=5,7$ and $13$.
More generally, for odd $q$ and $2n \ge 6$, ${\rm PSp}(2n,q)$ has the same order as but is not isomorphic to ${\rm P}\Omega(2n+1,q)$, and I would expect there to be many more resulting examples.
Of course it is very likely that there are examples coming from simple groups that do no have the same order  why not?

$\begingroup$ @ Derek Holt. I would prefer to restrict our simple groups to $A_{p}$,for some prime $p$, $PSL(p,q)$ and $PSU(p,q)$ for some prime $p$ and some prime power $q$. What about nwo? $\endgroup$ Jan 6, 2018 at 17:56