Number of ways to label die with opposing faces summing to 7 A friend and I have a disagreement about the number of unique ways to label the faces of a six-sided die with integers from one to six so that each pair of opposing faces sums to seven.
I think the answer is 8 since swapping each pair of opposing faces results in (I think) a different handedness for the two hemi cubes centered on the two faces being swapped, without affecting the other hemi cubes. So I think the problem is represented by the possible independent states of the three opposing pairs, giving $2^3=8$ permutations.
My friend thinks the answer is 2, as he argues that a single swap changes the handedness of the entire die, and that further independent swaps can’t further change the die’s handedness, and thus don’t generate unique states.
Who is right? (Or is neither of us right?)

Note that the faces with labels 1, 2, and 3 must appear in a 3-cycle (share a corner), and labeling these three faces uniquely determines the labels of the other three faces.  There are two ways to do that.
