How to determine whether a polytope is self-tessellating? I'm looking for polytopes that can be tessellated by a finite number of scaled versions of themselves. I'll use the term component for such a scaled version in the text below. 
Self-tessellating polytopes could be interpreted as the generalisation of rep-tiles and irreptiles to higher dimensions.
If we assume all components to be equal in size, irreptiles (and their higher-dimensional equivalents) are no longer relevant. As a result, all components should be congruent to one another (and, of course, similar to the original polytope). 
It would be very useful to have a general procedure to determine whether a given polytope is self-tessellating or not. Although it can be relatively straightforward to check or verify this property for some basic 2D or 3D polytopes, such as the cube —

which, incidentally, might look vaguely familiar —

more complex polygons and polyhedra or higher dimensional polytopes in general require a more structured approach. 
To get started, it might help to propose a couple of candidates. What about the following convex polytopes, are they self-tessellating or not?


*

*Tetrahedron

*Rhombic dodecahedron. Quite a few interesting properties (it is a zonotope and it's the "building brick" of the rhombic dodecahedral honeycomb)

*16-cell (4D octahedron), building brick of another honeycomb
 A: A self-tessellating figure can be dissected into some number of figures
each similar to the original figure.
I will generally assume that the smaller figures are all equal in size.
I find it easier sometimes to consider the dissection of a larger figure
into smaller ones rather than try to build a large figure from many
copies of a small one.
The regular tetrahedron
In order to self-tessellate the tetrahedron, 
one must dissect an edge of the tetrahedron into $n$ congruent line segments such that each of these segments is an edge of one of the tetrahedral pieces of the dissection of the original tetrahedron.
Now consider what happens when we remove these smaller tetrahedra from
the larger tetrahedron. The resulting figure has vertices that are
similar to the vertices of a regular octahedron. Consider one such vertex, $V$.
The dissection of the large tetrahedron would have to include one tetrahedron
with a vertex at $V$, but removing this tetrahedron leaves a figure that
clearly cannot be dissected into other tetrahedra of the same size.
Even if you remove the same-size requirement, as long as the number of
tetrahedra must be finite the dissection would still be impossible due to
the need to dissect a figure with a vertex like a regular octahedron.
An irregular tetrahedron
At least part of the argument about the regular tetrahedron applies to an
irregular tetrahedron. At least along edges whose dihedral angle is less than
twice the smallest dihedral of the tetrahedron,
one must arrange the same number of smaller tetrahedra along each
edge of the large tetrahedra, and the remaining figure when we remove
the small tetrahedra has vertices where four faces meet.
I think that in at least one such case it is impossible to fit one
tetrahedron to that vertex such that the remaining figure can still
be dissected, but this is harder to show than for the regular tetrahedron.
The rhombic dodecahedron
As with the regular tetrahedron, each edge would have to be the union
of the edges of smaller rhombic dodecahedra into which the larger dodecahedron
is dissected. But looking at just one small dodecahedron
with an edge coinciding with an edge of the larger dodecahedron,
the vertex between three faces of the smaller dodecahedron is also
the vertex between three faces of the figure remaining when we remove
the smaller dodecahedron, and that vertex is too "sharp";
there is no way to place a vertex of even one rhombic dodecahedron
at that vertex while still remaining inside the larger dodecahedron.
The regular $16$-cell polytope in four dimensions
Suppose we have a $16$-cell polytope with vertices at
$(1,0,0,0),$ $(-1,0,0,0),$ $(0,1,0,0),$ $(0,-1,0,0),$ 
$(0,0,1,0),$ $(0,0,-1,0),$ $(0,0,0,1),$ and $(0,0,0,-1).$
Consider the hyperface of this $16$-cell with all non-negative coordinates;
this is the regular tetrahedron $A$ with coordinates 
$(1,0,0,0),$ $(0,1,0,0),$ $(0,0,1,0),$ and $(0,0,0,1).$
If we are to dissect the $16$-cell into some number of
smaller $16$-cell polytopes, all the same size as each other,
some number of those small $16$-cells must have hyperfaces on $A$,
and those hyperfaces must completely cover $A$.
But all hyperfaces of a regular $16$-cell polytope are regular tetrahedra,
and we already know it is impossible to dissect a regular tetrahedron
into regular tetrahedra.
So it is impossible to dissect a regular $16$-cell polytope
into smaller $16$-cell polytopes.
By the way, the Schläfli symbol for the $16$-cell honeycomb is
$\{3,3,4,3\},$ indicating that three $16$-cells are joined at one
shared triangular face; hence the "dihedral" angle between any two
of those $16$-cells is $120$ degrees, and there is no way to slice
through the honeycomb along a single hyperplane without cutting
through some of the $16$-cells. This helps explain why the
$16$-cell honeycomb does not help us self-tessellate the 
regular $16$-cell polytope.
Other convex polytopes of more than two dimensions
Having shown that (sadly) none of the proposed figures is self-tessellating,
let's look now for figures that are self-tessellating.
Clearly, the hypercube (also called measure polytope) in any number of
dimensions is self-tessellating. So is any linear transformation
of the hypercube.
In any dimension greater than two there are
prisms other than the hypercube (and its linear transformations)
that are self-tessellating.
We can construct some of these as follows:
Take any self-tessellating plane figure $A$ and an orthogonal projection of $A$ onto another plane parallel to the plane of $A$.
Dissect $A$ into $n^2$ congruent pieces similar to itself,
where $n$ is an integer. (If you have a dissection into $m$ pieces where
$m$ is not a perfect square, simply perform a similar dissection of each piece
of the first dissection so you have $m^2$ congruent pieces and set $n=m$.)
Then you can cut the prism into $n$ equal "slices" along planes parallel to 
the face $A$, and then use the dissection of $A$ to dissect each "slice"
into $n^2$ congruent pieces each similar to the original prism;
then you have a self-tessellation of a prism by $n^3$ smaller congruent prisms
similar to the first prism.
To create such objects in higher dimensions, we start with a
$d$-dimensional object dissected into $n^d$ congruent pieces similar to itself.
We then project this figure onto a $d$-dimensional hyperplane parallel
to the hyperplane of the original object, and cut the resulting
$(d+1)$-dimensional "prism" into $n$ equal "slices" along parallel hyperplanes.
This gives a dissection of the $(d+1)$-dimensional "prism" into
$n^{d+1}$ congruent figures similar to itself.
Non-convex polytopes of more than two dimensions
Self-tessellating non-convex prisms can be made with self-tessellating
non-convex polygons as their end faces.
For example, consider this dissection of the bent triomino 
into four congruent pieces:

Now suppose that we cover this figure with four copies of the bent tricube
(one of the pieces of the Soma puzzle;
also described here
and here):


If we stack another layer of tricubes on top of these four in the same pattern,
the result is a dissection of a larger tricube into $8$ smaller ones.
Consider the L-shaped and T-shaped tetracubes:
    

The L tetromino can be dissected into four smaller copies of itself,
so the L tetracube can be self-tessellated by eight smaller L tetracubes.
The T tetromino  cannot be dissected into four T tetrominoes,
but a $4\times4$ square can be dissected into four T tetrominoes;
hence a tetromino (which consists of four squares)
can be dissected into $16$ tetrominoes.
Cover this dissection with $16$ T tetracubes and stack four such layers
atop each other; then you have a dissection of a T tetracube 
into $64$ smaller congruent T tetracubes.
Now consider these three Soma pieces:
    
    

A cube can be dissected into two congruent copies of any one of these figures.
Since each figure is made of four cubes, each of them is self-tessellating
by eight smaller congruent shapes similar to itself.
This shows that there are three-dimensional shapes other than prisms
that are self-tessellating.
