Is the rhombic dodecahedron the only isohedral polyhedron that tiles 3-space (other than the cube)? Is the rhombic dodecahedron

the only face-transitive (or isohedral, i.e. all faces are the same) polyhedron that seamlessly tiles 3-dimensional Euclidean space (other than the cube)?

I'm looking for an answer to this question and although Wikipedia provides a lot of lists for 3-tesselations I cannot find a definite closure. In particular, if the above statement were true I'd expect it to be listed on the tesselation's Wiki page, but no such statement exists, which leaves some doubt in me whether it is actually the case.
 A: No. The body-centered cubic tetrahedron tiles it as well. 
Please note that an isohedral polyhedron is not just a polyhedron in which all faces are congruent, but one in which all the faces also lie in the same symmetry orbit (which unfortunately is not the case for the irregular space-filling octahedron).
A: Does the (irregular) space-filling octahedron meet your criteria? (The tiles are not all obtained from a single tile by translation, i.e., this is not a lattice tiling. Instead, there are three families of mutually-orthogonal tiles in a tessellation.)

A: The rhombohedron works, though you may consider it too closely related to the cube to be of any interest:

A: The rhombic dodecahedron could be split-up into six isohedral octahedrons mentioned above.
But each octahedron could be split-up further into four isohedral tetrahedrons (so the rhombic dodecahedron consists of 24 isohedral tetrahedrons).
Below is the view of this isohedral tetrahedron inside the half of the cube:

A: Truncated octahedron also tesselates 3-dimensional space.
