I think there is a general belief that the classification of all finite groups is "impossible". I would like to know if this claim can be made more precise in any way. For instance, if there is a subproblem of the classification problem that is already equivalent to an already agreed-upon wild problem. 



In what sense is the classification of all finite groups "impossible"?
One can make the argument by wildness much more concrete than in the previous answer: Sergeichuk ["Classification of metabelian p-groups", in: Matrix problems, Inst. Mat. Ukrain. Akad. Nauk, Kiev, 1977, pp. 150-161, in Russian] showed that isomorphism of 2-step nilpotent p-groups is already wild (over $\mathbb{F}\_p$), whenever the center is not cyclic of order p.


To answer David Harden's question from the comments, which also gets at Timothy Chow's point about computational complexity: simultaneous conjugacy of k-tuples of matrices can be solved in polynomial time [[Sergeichuk](http://www.ams.org/mathscinet-getitem?mr=1782204); [Brooksbank-Luks](http://www.ams.org/mathscinet-getitem?mr=2464805); [Chistov-Ivanyos-Karpinski](http://www.ams.org/mathscinet-getitem?mr=1809971)] (or, I believe, even in $\mathsf{NC}$, depending on the field and model of computation; an even simpler algorithm puts it in $\mathsf{RNC}$). However, the problem of isomorphism of 2-step nilpotent $p$-groups is Tensor Isomorphism-complete [G-Qiao](https://arxiv.org/abs/1907.00309). Another TI-complete problem is conjugacy of $k$-dimensional *subspaces* of matrices. [Belitskii and Sergeichuk](http://www.ams.org/mathscinet-getitem?mr=1955562) showed that the latter classification problem is strictly harder than $k$-tuple conjugacy (that is, subspace conjugacy contains k-tuple conjugacy but not conversely). When the subspaces are given by a spanning set, subspace conjugacy [is at least as hard as Graph Isomorphism](http://www.ams.org/mathscinet-getitem?mr=3026328) [Chapter 4 [here](http://tuvalu.santafe.edu/~jgrochow/grochow-thesis.pdf) contains a freely available and more complete version] (that is, Graph Iso reduces to subspace conjugacy in polynomial time), which is not known to be in $\mathsf{P}$. It is even [harder than Code Equivalence](https://dx.doi.org/10.4230/LIPIcs.ITCS.2021.31) (which itself is thought to be harder than Graph Iso).


