I was watching “Who cares about topology? (the inscribed rectangle problem)” by $3$Blue$1$Brown, and it explained how the unit square, when considered as representing all unordered $2$-tuplets, maps to the Mobius Strip.
I noticed that it was folded in half because it has one axis of symmetry, with two equivalent sides, which is equal to the number of permutations of a two-element set.
When we consider the unit cube, it has three sides, or three coordinates. This gives $3!$, or $6$ permutations per set, or $6$ axes of symmetry. This means that $6$ volumes within the cube can be considered equivalent if we consider the points as unordered, instead of ordered, sets.
Now, in the case of the directed unit square as derived from sets of two points around a circle, to preserve the manifold structure, one side had to be given a twist (thus resulting in a Mobius Strip).
Take the unit cube, as derived from the set of unordered triples between zero and one, which correspond to points around a circle. If we were to somehow bend each (in our case, considered equivalent) volume in the fourth dimension so that all faces of these volumes formed a different, contiguous and continuous volume of some kind, what would we create?
Essentially, what volume (closed manifold, I think) does this particular unit cube map to?
Edit: It has been brought to my attention that, in the original context, the coordinates of the points corresponded to individual numbers on a circle, (basically, distributed across the circumference of a circle and identified with a single number) and were then mapped onto the unit square. As such, my question really regards the unit cube as derived from all unordered triples of points around that circle. The question has been modified to match this.