Polyhedra having equal quadrilateral faces are cubes?

Question

While discussing with my 11 y.o. daughter about the definition of a cube as regular hexahedron, I observed that actually we can let drop the assumption that the faces are squares, and require only that they are equal. She agreed, because trying to imagine, let's say, six equal parallelograms (without right angles) matching to form a solid leads to impossible solids.

I recently made a conjecture requiring even weaker conditions.

A polyhedron having equal quadrilateral faces is a regular hexahedron.

Any hint to prove the statement or a couterexample would be great

Thank you in advance for your attention

Answer 1

Consider the image of the cube under the map $(x, y, z) \mapsto (x+(x+y+z), y + (x + y + z), z + (x + y + z)$. The faces are all still quads, and all still congruent (by symmetry), but it's not a cube.

Answer 2

You can "bookcase squash" a cube in two directions, making the faces congruent rhombi, and get a Rhombohedron

Answer 3

Note that the quadrilaterals do not have to be rhombuses, either. The deltoidal icositetrahedron has 24 identical quadrilateral faces, each of which is a "kite" (two pairs of adjacent edges equal, but not all four edges equal).