This is the 3D (and higher D) version of https://mathoverflow.net/questions/370302/a-claim-on-partitioning-a-convex-planar-region-into-congruent-pieces

- Is there a 3D convex polyhedral solid that can be cut into 2 mutually congruent non-convex polyhedral solids but *not* into 2 mutually congruent convex solids? An example with say, 3 replacing 2 will also do.

Remark: I had thought that Noam Elkies's pentagon on the above linked page (it allows partition into 2 non-convex and congruent polygons but not into 2 convex congruent polygons) can be 'lofted' into a 
 pentagonal pyramid - with every horizontal section a scaled down copy of the pentagonal base - that can be cut into 2 congruent non-convex polyhedrons but not into 2 convex and congruent polyhedrons. But it doesn't look obvious that congruence of the 2 pieces can be achieved naturally. 

If a polyhedron with the desired property is found, one could try to *characterize* all polyhedrons that can be cut into 2 congruent non-convex polyhedrons but not into convex congruent polyhedrons. Increasing the number of pieces in the question  might make the question more difficult.

**Note:** As has been noted more recently in the above linked page, even on the plane, I don't know any convex *polygon* with even number of sides that has the desired property.