Can a group be fully characterised by its Sylow $p$-subgroups?

Question

I was wondering if we have two groups, $G$, $H$, such that $|G| = |H|$ (and these orders are finite); and all their Sylow $p$-subgroups have the same structure (i.e. Syl$_p(G)$ isomorphic to Syl$_p(H)$ for all $p$) and each Sylow $p$-subgroup has an equal number of conjugates in $G$ and $H$, then can it ever be such that $G$ is not isomorphic to $H$?

My instinct on this is no, though it is hard to find examples of relatively small groups where it would be applicable. I asked the Group Theory lecturer in my maths department and they suspected there are such $G$ and $H$, but did not give an example and only cited that "group classification would be much easier if so", so I was wondering if somebody here could provide either a proof that $G$ and $H$ must be isomorphic, or a counterexample?

Answer 1

A counterexample is $\mathtt {SmallGroup}(147,4)$ and $\mathtt {SmallGroup}(147,5)$. They both have elementary abelian Sylow $7$-subgroups and self-normalizing Sylow $3$-subgroups.

Both groups have the structure $7^2:3$, i.e. semidirect products $C_7^2 \rtimes C_3$, but the action of $C_3$ is defined by a scalar matrix in the first group, and a non-scalar diagonal matrix in the second group.

They can be defined by presentations $$\langle x,y,z \mid x^7=y^7=z^3=1, z^{-1}xz=x^2,z^{-1}yz=y^2\rangle$$ and $$\langle x,y,z \mid x^7=y^7=z^3=1, z^{-1}xz=x^2,z^{-1}yz=y^4\rangle.$$ The first one has a larger automorphism group than the second. I don't know whether there is an easier way to prove that they are not isomorphic.

Edit: As verret pointed out in a comment, an easier way to see that the groups are not isomorphic is that in the first group all eight subgroups of order $7$ are normal in $G$, whereas in the second group the only normal subgroups of order $7$ are $\langle x \rangle$ and $\langle y \rangle$.

Another Edit: The groups $\mathtt {SmallGroup}(605,5)$ and $\mathtt {SmallGroup}(605,6)$ with structure $11^2:5$ are even more alike. They have the stipulated properties, and they both have exactly two normal subgroups of order $11$.