Htykuut

If $G$ has that $|G| = p^am$ where $p$ is prime and $gcd(p,m) = 1$, we have that $n_p(G)$ counts of Sylow $p$-subgroups in $G$. If $P$ is a Sylow $p$-subgroup, by the third Sylow theorem
begin{align*}
n_p(G) &equiv 1text{ mod } p\
n_p(G) &| m\
n_p(G) &= [G:N_G(P)]
end{align*}
$n_p(G)$ seems to depend on $G$, and these properties seem to only give possible candidates for what $n_p(G)$ will be. If I am given $G$ and $H$ with $|G| = |H| = p^am$, do we necessarily have that $n_p(G) = n_p(H)$? It feels sketchy to say so, but I can't come up with any counterexamples off the top of my head.

Hint: take $S_3$ and $C_6$ Both groups of order $6$. And look at the prime $2$.

By the way, suppose you have two finite groups $G$ and $H$ of the same order and for every prime $p$ dividing the order of $G$, $n_p(G)=n_p(H)$. Then $G$ and $H$ do not have to be isomorphic! (Example, take two non-isomorphic (non-abelian) groups $P_1$ and $P_2$ of order $p^3$ (see here), $p$ an odd prime and look at $G=C_2 times P_1$ and $H=C_2 times P_2$)