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.
48 ! :-)
(special case)


  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.


It is clear from both your initial disagreement and comments and answers here that the question is somewhat ambiguous - the meaning of "unique ways" needs refining. eg does 'handedness' matter, or .... . 

So, based on what's been said since, while it's clearly not what you originally intended, here is an interpretation that seems valid (to me :-) ) that yields the result 48!
This is based on the assumption that the unnumbered die is NOT rotation agnostic but itself has uniquely identifiable faces apart from the numbering. While that's not how die are usually made, it's a valid (I think) variation on the original question. 

Imagine a cube with faces inlaid with 6 different semi-precious stones. (It could equally well have 6 different colours but Semi precious stones seems more interesting.) If you have multiple copies of a standard die with colours assigned in the alignment specified below, and then assign valid numbers to faces and produce a set of varying dice, how many unique dice can you create?

Assume a standard SPS die, with say topaz up, chalcedony down, beryl left, jasper right,  carnelian front and amethyst back. If we decide this is going to bend the brain we can resort to calling the sides Up, Down, Left, Right, Front, Back - or even U D L R F B.

Start with Up / Topaz.
We have 6 numbers to choose from.
Assigning any number also sets the number of Down/Chalcedony to make the opposing faces sum to 7.    

Now assign a number to say Left / Beryl
We have 4 choices - and this also sets the value of Right / Jasper.  

That leaves 2 choices for Front / Carmelian
and sets Back / Amethyst accordingly.

Total unique die are 6 x 4 x 2 = 48.



But wait, there's more!
Maybe.  

Here's an arguably* new question arising which I may propose as such.
But, it's worth outlining here.
*"Arguably" as this too seems to be a legitimate interpretation of the original question. 
Just as I randomly assigned numbers to face-pairs, I could assign stone types to faces. But as there is no sum-to-7 constraint there are 6! assignable patterns.
However, many of these are non unique if the unnumbered cube is rotated.
How many unique numbered cubes can we make if the cubes can be assigned stones to faces randomly before numbering?

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The special collectors edition 

uses a Sapphire, Emerald, Sardonyx, Chrysolite, Chrysoprase &amp; Jacinth die. 

Bonus question -  where did I get my two stones lists from?