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Saúl RM
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Keller's cubing conjecture but with arbitrary cubes of side $1$

These days I have been reading about Keller's cube tyling conjecture, which asks if in any covering of $\mathbb{R}^n$ by translates of $[0,1]^n$ with disjoint interiors there are two cubes sharing one $n-1$-dimensional face (the problem was completely solved recently in The Resolution of Keller’s Conjecture, by Brakensiek, Heule, Mackey and Narváez).

I have not found any information on the same problem but without assuming that the cubes are translates of each other, that is, for any covering of $\mathbb{R}^n$ by n-dimensional hypercubes of side one (sets isometric to $[0,1]^n$) with disjoint interiors, can we always find two hypercubes sharing one face? The interesting case is $n\leq7$, for $n\geq8$ there are counterexamples to the original conjecture.

There are coverings of $\mathbb{R}^3$ by disjoint cubes such that they are not all translates of each other, so at least this is not a vacuous question.

Saúl RM
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