The broad theme that underlies this question is: to what extent can the study of finite groups be reduced to the study of $p$-groups?
I imagine that it is possible for a pair of nonisomorphic finite groups $G$ and $H$ to have isomorphic Sylow subgroups. That is to say, for every prime $p$, $G$ and $H$ have isomorphic Sylow $p$-subgroups. In particular, $G$ and $H$ must have the same number of elements.
Let us call $G$ and $H$ Sylow-isomorphic if they have isomorphic Sylow subgroups.
For a given positive integer $n$ (that we may obviously assume isn't a prime power), how many Sylow-isomorphic finite groups of order $n$ can there be?
Can a pair of non-isomorphic finite simple groups be Sylow-isomorphic?
Some results:
Every finite nilpotent group is isomorphic to the external direct product of its Sylow subgroups, so a pair of finite nilpotent groups are Sylow-isomorphic if and only if they are actually isomorphic.
Let $C_n$ denote the cyclic group of order $n$. If $p>2$ is an odd prime, $\text{Aut}(C_p)\cong (\mathbb{Z}/p\mathbb{Z})^\times\cong C_{p-1}$. We know that $C_p\rtimes C_{p-1}$ and $C_p\times C_{p-1}$ are non-isomorphic (the former is non-abelian while the latter is abelian), but clearly they are Sylow isomorphic. This shows that it is possible for a pair of non-isomorphic solvable groups to be Sylow-isomorphic.
Additional question:
I've heard that it is often useful to study the normalizers of $p$-groups in finite groups (this method is sometimes called local analysis, and normalizers of $p$-groups are called $p$-local subgroups).
Since the Sylow $p$-subgroups of a finite group $G$ are conjugate, their normalizers must also be conjugate, so we can unambiguously talk about the isomorphism type of a Sylow $p$-normalizer of a finite group. Suppose $G$ and $H$ have isomorphic Sylow $p$-normalizers for every prime $p$. This also means, by the way, that $G$ and $H$ must be Sylow-isomorphic. In this situation, let's call $G$ and $H$ locally isomorphic. Then what can we say about how $G$ and $H$ are related?