What is the hardest instance for the group isomorphism problem?
$p$-groups of class 2 and exponent $p$ are widely believed to be the hardest case of Group Isomorphism ($p &gt; 2$). (For $p=2$, we need to consider exponent 4, since all groups of exponent 2 are abelian - easy exercise for the reader.) Although there is as yet no reduction from general GpIso to this class of groups (though see point 0.5 below), there are several reasons for this belief. Let me outline some of them here.

0) Practical experience (see papers by Newman, Eick, O'Brien, Holt, Cannon, Wilson, ... which give the algorithms that are implemented in GAP and MAGMA).

0.5) [EDIT: added 8/7/19] Reductions. When such $p$-groups are given by generating sets of matrices over $\mathbb{F}_p$, the problem is $\mathsf{TI}$-complete [G.-Qiao '19]. Also (cf. point (4) below), isomorphism of $p$-groups of exponent $p$ and class $c &lt; p$ reduces in poly time to isomorphism of $p$-groups of exponent $p$ and class 2 (ibid.).

1) Structure (reduce to solvable, then to $p$-group). Every finite group contains a unique maximal solvable normal subgroup, called the solvable radical, denoted $Rad(G)$. $G/Rad(G)$ contains no abelian normal subgroups, and isomorphism of such groups can be handled efficiently in practice (Cannon-Holt J. Symb. Comput. 2003) and in theory (Babai-Codenotti-Qiao ICALP 2012). Even for groups where $Rad(G)$ is abelian, some of these can be handled in $n^{O(\log \log n)}$ time (G-Qiao CCC '14, SICOMP '17) - so, not quite polynomial, but much closer than $n^{\log n}$. The main obstacle thus appears to be solvable (normal sub)groups. Now, within solvable groups, there is a lot of structure - starting with the fact that every solvable group is a knit product of its Sylow $p$-subgroups - and it seems the hardest cases are $p$-groups.

2) Counting. The number of groups of order $n$ is $\leq n^{(\frac{2}{27} + o(1))\mu(n)^2}$, where $\mu(n)$ is the largest exponent of any prime dividing $n$ (Pyber 1993). The number of $p$-groups of order $n=p^m$ is at least $p^{(\frac{2}{27} + o(1))m^2}$ (Higman 1960). So you see that the coefficient of the leading terms in the exponents match. In this sense "most" groups are $p$-groups (even of class 2 and exponent $p$). There is a long-standing conjecture which says that "most" in the preceding weak sense can be strengthened to say that proportion of groups of order $\leq n$ which are $p$-groups tends to 1 as $n \to \infty$.

3) Universality (/wildness). Giving a classification of $p$-groups would imply a classification of all modular representations of any finite group (or even Artinian algebra) in characteristic $p$ (Sergeichuk 1977). 

4) Flexibility. Why $p$-groups of class 2 and not higher class? (Note that $p$-groups of nearly-maximal class, so-called "small coclass", have essentially been classified, Eick &amp; Leedham-Green 2006, see also some of the answers here.) To any $p$-group one can associate a graded Lie ring, where bracket in the Lie ring corresponds to commutator in the group. Associativity in the group implies the Jacobi identity for the bracket, thus giving rise to a genuine Lie ring. However, note that when the group is class 2, the Jacobi identity is trivially satisfied (all its terms are automatically 0), so this puts no additional constraints on the structure. It basically just corresponds to an arbitrary skew-symmetric bilinear map. For $p$-groups of exponent $p$, there is even a reduction from class $c &lt; p$ to class 2.