Someone else remarks that it's "got to be all edges flipped in place",
and Jerry Bryan remarks that it is.

   *6*              *6*
   6*6              3*4
   *6*              *1*
   *2*              *5*
   2*2              3*4
   *2*              *2*
*3**1**4*        *1**1**1*
3*31*14*4        5*23*42*5
*3**1**4*        *6**6**6*
   *5*              *2*
   5*5              3*4
   *5*              *5*

I disagree. Look at the 1-2 edge. If it's "flipped in place", then
since it appears to be fixed, the cube must flip around it. But then
the four 3 faces would be where the 4 faces actually are. No, it's
more complicated than just all-edges-flipped.

"[Q]uite extraordinary and wonderful" it is.

It is in fact the position arrived at by flipping all edges in place, *then*
reflecting the entire configuration. I believe this also tells us what the
other two equivalence classes with just 24 elements are: they are the
results of doing each of these two operations separately.

David Seal
dseal@armltd.co.uk