(Sorry.) The right answer should be:

The state of the cube is not:

X|O|X|X|X                  X|A|X|C|X
X|X|X|X|X                  X|X|X|X|X
X|X|X|X|X     But rather:  X|X|X|X|X
X|X|X|X|X                  X|X|X|X|X
X|X|X|O|X                  X|X|X|B|X

Where cubie "C" just "looks" like it's in the right place.

You need an operator that rotates A->B->C->A. (Left as an exercise; hints
available on request.)

This will very likely leave an inconvenient number of edges flipped. For
the answer to _this_ problem, see my last post. ;)

I think you must be wrong here (but would love to be proved wrong--I'm no
mathematician so group theory is very much beyond me).

Proof #1:
We hold the cube with the red face on top, and the yellow face in
front (colors obviously don't matter, but I find it easier to discuss using
them). We will assign a parity to the edge cubies, being defined by holding
the cube such that the red face of the cubie is on top and the yellow in
front. If the cubie is on the left as we look at it in this position it is
parity 0, on the right it is parity 1.

There are only two operations available which affect the cubie we
are interested in: rotating the front face 90deg; and rotating the slice the
cubie is in 90deg. It is easy to see that neither of these moves alters the
parity (assume the cubie's frame of reference, and think of rotating the
rest of the cube around it--it is clear that it will not end up on the other
side).

Therefore the move C->A in the above is impossible.

Proof #2:
Take apart the order 4 cube (my falls apart depressingly easilly)
and try to reassemble it with the two edges exchanged. It will not fit, as
they are mirror images of each other.

Note that you get an apparant parity reversal by flipping the
cubies, but this does not actually move anything. In other words, no amount
of flipping and moving will allow you to end up moving A->B->C->A. That's
why I solve edges first.

Ronnie