This is the 3D (and higher D) version of 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 can further ask how 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.