$3\mathrm D$ Projection Of $4\mathrm D$ Polyhedron 
Can someone identify this shape?


I think it is a $3\mathrm D$ projection of $4\mathrm D$ polyhedron. The body in the center seems to be a truncated octahedron, so as the body in the middle. The outer one is a snub cube, I think.


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*Can someone help me to determine   whether it is indeed a $3\mathrm D$ projecttion of a $4\mathrm D$ polyhedra? 

*If it is, what is the $4\mathrm D$ polyhedron and what is the projection centered at? 

 A: As I start writing, I'm not convinced it's even a $4$-polytope. It probably is, but I'll at least attempt a decomposition into $3$-faces and hopefully compute the Euler characteristic. I don't believe it's a well-known polytope, and I think the 'outer' facet is actually a truncated octahedron with hexagonal pyramids attached to each square face (there are several vertices of degree $6$, and the snub cube doesn't have any of those).
I definitely believe the inner facet is a truncated octahedron (I'll call any of these $P^3$, for the $3$-dimensional permutohedron; also I keep writing 'icosahedron' instead of 'octahedron'). Also, please excuse the right side of my highlighted edges, something seems a little off.

Now, when it comes to those "joining" facets, things are a little more complicated, but not too much so. Essentially, we almost have prisms coming off each face of the central $P^3$, but not quite. For the square faces of $P^3$, we have

two cubes, depicted here with red and blue edges, sharing the purple square $2$-face. Let's start counting faces of each dimension now.
We get $12$ vertices for each of the $6$ square faces of the central $P^3$, for a total of $72$ with only a few missing. 
Each cube duo contributes $5 \cdot 4 = 20$ edges, for $120$ so far. 
Assuming the purple squares are legitimate $2$-faces (I'm not sold on this), we have $6 + 5 = 11$ faces of dimension $2$, for $66$ so far, and finally $2 \cdot 6$ facets, plus the central $P^3$, for $13$ facets.
The final sort of facets are a bit odd

as they seem to be an 'inner' hexagonal prism (red edges) and a hexagonal prism with hexagonal pyramid 'attached' at a hexagonal $2$-face (blue edges), with each of these facets also sharing a purple hexagonal $2$-face.
There are $8$ of these facet pairs (one for each hexagon in the central $P^3$), bringing our total up to 29 (plus the outer facet), for $30$ total facets.
These facet pairs contribute only $1$ vertex not yet accounted for, and we have $72 + 8 = 80$ vertices by my count.
Now, many of the edges of these facets are contained in things listed so far. It'll be helpful to consider the $12$ green edges of the central $P^3$, each will give us $3$ edges yet uncounted, plus $6$ edges (from the apex of the pyramid) for each of the $8$ hexagonal faces of the central $P^3$. This is $12 \cdot 3 + 6 \cdot 8 = 84$ new edges, for a total whopping count of $120 + 84 = 204$ edges. 
Finally, for $2$-faces, each of the $12$ green edges of the central $P^3$ gives $2$ new square $2$-faces. Each of the $8$ hexagonal faces of the central $P^3$ gives $2$ hexagonal $2$-faces (one in the central $P^3$, one with purple edges in the picture) and $6$ triangular $2$-faces, for a total of $12\cdot 2 + 8\cdot 8 = 88$ new $2$-faces, and a grand total of $66 + 88 = 154$ faces of dimension $2$.
Thus, it seems like the $f$-vector is $(80, 204, 154, 30)$, and the alleged Euler characteristic is indeed $80 - 204 + 154 - 30 = 0$, as it should be for any $4$-polytope.
