Space-tiling convex prisms A convex prism is a subset of $\mathbb{R}^3$ congruent to the Cartesian product of a convex polygon (the prism's base) with the interval $[0,1]$.

Question. If a family of congruent convex prisms tiles space (not necessarily in a face-to-face manner), must there exist a tiling of
  the plane with polygons congruent to the prism's base?

 A: [The following is not quite an answer, but it refutes a natural
generalization suggested in the Comments, and is too long to be
a comment itself.]
Counterexample in ${\bf R}^N \times {\bf R}$ for some $N>2$:
any lattice hexagon $H$ with angles 
$90^\circ$, $90^\circ$, $135^\circ$, $135^\circ$, $135^\circ$, $135^\circ$ 
in that order.  (That is, $H$ is obtained from a lattice rectangle by
truncating two adjacent vertices by two isosceles right lattice triangles, 
not necessarily congruent.  Alternatively, remove two congruent
lattice triangles, not necessarily isoceles, related by a 90-degree rotation.)
Then $H$ does not tile the plane, but
four copies do tile a (non-simply-connected) polyomino.  But it was
recently shown that any polyomino in some ${\bf Z}^n$
(which need not be simply connected, or even connected at all!) 
tiles ${\bf Z}^d$ for some $d$:

Vytautas Gruslys, Imre Leader, and Ta Sheng Tan: Tiling with arbitrary tiles.
  Proc. London Math. Soc. (2016) 112 (6): 1019-1039.
  https://doi.org/10.1112/plms/pdw017 $\cong$ http://arxiv.org/abs/1505.03697

(I learned about this from Francisco Santos's accepted answer to 
Timothy Chow's 
 Mathoverflow question 49915, which the MO algorithm helpfully put at 
the top of its list of questions "Related" to this one.)
