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There seems to be some divisibility by h/2, where h is the Coxeter number. The Auslander-Reiten translation acts on the set of bases for the K-group. The order of every orbit divides h. Not clear to me what the stabilizers can be.
There is a necessary condition, that the Coxeter polynomial is a tensor product of two polynomials. This can be checked for instance on the set of roots.
In singularity theory, one can make the simple singularity E6 ($x^3+y^4$) from A2 ($x^3$) and A3 ($y^4$). This sum-with-disjoint-variables is named the Thom–Sebastiani sum.
For the next permutohedron, the squares are no longer facets, so one cannot just remove the squares by removing the associated inequalities defining the polytope, as was the case in dimension 3.