It seems that there are convex polyhedra which tile 3D space, for which any such tiling must have a vertex of some tile lying on the interior of a face of another tile.

Consider the polyhedron which is the convex hull of the following points:

$$(0,0,4), (2,0,2), (2,0,-2), (0,0,-4), (-2,0,-2), (-2,0,2), (3,\sqrt{3},0), (-3,\sqrt{3},0), (0,2\sqrt{3},2), (0,2\sqrt{3},-2)$$

This polyhedron can tile an unbounded hexagonal prism: Place three copies of the tile by rotating the shape defined above by 0, 120, and 240 degrees around the edge through $(0,2\sqrt{3},2)$ and $(0,2\sqrt{3},-2)$. Then take a copy of those three tiles, rotate them by 60 degrees around the same line and shift by 4 in the $z$ direction. These six tiles can then be repeated indefinitely by shifting by 8 in the $z$ direction.

So far, everything is vertex-to-vertex. But notice that there are regularly spaced vertices lying in the middle of each side of each hexagonal prism, and the location of these are offset between adjacent sides of the prism. We can put two hexagonal prisms together aligning these points vertex-to-vertex, but then the prism which touches those two cannot be placed to have all the vertices line up.

Of course, I have not proven that there is not some other way to tile 3D space with this tile. But I suspect that the only tilings with this tile involve forming hexagonal prisms, in which case the result holds. Note also that the polyhedron I described can be stretched by an arbitrary factor in the $z$ direction to form a family of examples of this form, in case the parameters I chose above happen to admit some surprising alternate tiling.

non-prismaticpolyhedrons that pack perfectly and have this property (Prisms are 2D in some sense, aren't they?)? And one still suspects that there probably are no convex polyhedrons that are perfect packers only when some vertices touch an interior point of a face of another unit. $\endgroup$