Same number or each Sylow p-subgroup $\implies$ isomorphism?

Question

I have a question about Sylow theory and isomorphic groups. I came with it while trying to solve some problems from my group theory course. The question is if this statement is true in general:

Being $G$, $H$ two groups of same order, $|G|=|H|=p_1^{\alpha_1}\cdots > p_n^{\alpha_n}$ with each $p_i$ prime. Then, if the number of Sylow $p_i$-subgroups in $G$ equals the number in $H$ for all $i=1,...,n$, then $G\cong H$.

Is this true? Is it true just in certain cases? If not, why? Any help will be appreciated, thanks in advance.

Answer 1

This looks false to me. Take $Q_8$ and $D_8$ for instance. Then, they have the same number of Sylow 2-subgroups (since their orders are $8$, they both have one) but they are not isomorphic.

Answer 2

Since there was a discussion under my previous answer, for consistency, I am adding this as a new answer for the cases where order of the group has more than one distinct prime factors:

Take $\mathbb{Z}_6\times \mathbb{Z}_2$ and $\mathbb{Z}_{12}$ for instance. Then, since they are abelian, Sylow 2-subgroups and Sylow 3-subgroups are normal in these groups. This means, $n_2=n_3=1$ for both but they are not isomorphic.