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Suppose that we have a polyhedron whose faces have been colored red and blue such that no two same-colored faces share an edge, and suppose that there are the same number of faces of each color.

Does it follow that the polyhedron is symmetric between its red and blue faces? In other words, is there necessarily a permutation of the nodes that maps each red face onto a blue face in an orientation-preserving (or reversing) manner?

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Of course not. That's a very weak condition. Basically, you've just equated the total number of vertices of red and blue faces. That's not nearly enough to enforce any symmetry at all.

Look at this guy: top view (left), bottom view (right).

polyhedron

See that? There is only one vertex of degree 8, and there are two quadrilateral faces sharing it. Good luck mapping them onto each other, considering that one of them shares the edge CD with yet another quadrilateral, while the other doesn't.

With relatively little effort, even more obvious examples can be built.

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