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?
Edit: on finding a pair of non-isomorphic, but locally isomorphic groups
If $G$ has a normal Sylow $p$-subgroup for any prime $p$, then any group that is locally isomorphic to $G$ must be isomorphic to $G$, so we must restrict our attention to groups without any normal Sylow subgroups. This means we must ignore supersolvable groups altogether, since if $G$ is supersolvable and $p$ is the largest prime divisor of $|G|$, then a Sylow $p$-subgroup of $G$ is normal.
Also, one might consider a group in which the notion of "locally-isomorphic" coincides with "Sylow-isomorphic": that is, groups in which every Sylow subgroup is self-normalizing. Such a group (assuming it isn't of prime power order) cannot be solvable, because of the following theorem:
Theorem: (R. Carter) Let $G$ be a finite solvable group. Then $G$ has a self-normalizing nilpotent subgroup, and all such subgroups are conjugate (and thus isomorphic)
So if $G$ is solvable, then there is at most one prime $p$ such that a Sylow $p$-subgroup of $G$ is self-normalizing.
Best Answer
To answer your question 2, there are very few pairs of finite simple groups with the same order. There's $PSL_3(4)$ and $PSL(4,2)$, and there's $P\Omega_{2n+1}(q)$ and $PSp_{2n}(q)$ for $q$ odd. None of these are Sylow isomorphic in your sense.
As a contribution to your question 1, finite groups can be not only Sylow isomorphic but they can even be indistinguishable from the point of view of fusion at any prime. This is equivalent to the $\mathbb{Z}$-completion of the classifying spaces being homotopy equivalent. Probably the smallest example is $(\mathbb{Z}/3\rtimes\mathbb{Z}/4) \times (\mathbb{Z}/5\rtimes\mathbb{Z}/2)$ and $(\mathbb{Z}/3\rtimes\mathbb{Z}/2) \times (\mathbb{Z}/5\rtimes\mathbb{Z}/4)$, with the action of $\mathbb{Z}/4$ having the subgroup of index two in its kernel in both cases.