Precisely. So the error arrived when we removed the extra points and IIA implied the decision didn't change.
What? (2,2) is still on the axis of symmetry, regardless of whether or not the point (2.6, 1.2) exists or not, and so if they select that point because of symmetry, they will continue to do so, regardless of the existence or nonexistence of irrelevant alternatives.
Scalings are aretefacts of how we represent the utility functions. They have no intrinsic meaning. So I have not been using a "symmetry-braking transformation" - I haven't used a transformation at all, just a different way of drawing the exact same situation.
Not in this problem, because you've set it up so the axis of symmetry is used for decision-making. If you do the scaling to both the points and the axis of symmetry, then you get the correct answer- (1.0, 0) - because it lies on the line y=2x-2, which is the new axis of symmetry. By only scaling part of the problem, you perform a transformation.
[edit] Thinking about this more, I think I've modified my position somewhat: you have two assumptions, first that the outcomes are symmetric and second that there are no canonical choices for utility functions. Those don't look like they play well together- if outcome (0,3) is symmetric with outcome (3,0), and then by changing your choice of utility function you can make (0,3) symmetric with (2,2), then you have a serious problem. If you fulfill the symmetry assumption by restricting IIA to removing pairs of symmetric points, the crisis is averted.
[edit2] There's a more general point that should be mentioned- whenever you have a decisionmaker whose decision depends on relative outcome merits, then IIA either breaks or is limited severely. "Irrelevant" needs to be understood not as "outcomes I didn't choose" but "outcomes which didn't impact my choice." When your rule is "pick the second best," then both the best and second best are relevant alternatives, even though you only picked one. In a bargaining game without a frame, the only way to judge outcomes is by their relative merits- and so you get weak or broken IIA.
But in this particular problem, the symmetry assumption throws a wrench into that general point, because now there is a frame- the symmetric lattice, and the implied axis of symmetry- and so there's an objective criterion rather than simply relative criteria.
What? (2,2) is still on the axis of symmetry, regardless of whether or not the point (2.6, 1.2) exists or not, and so if they select that point because of symmetry, they will continue to do so, regardless of the existence or nonexistence of irrelevant alternatives.
The axis of symmetry is a property of the figure (in this case, the set of points), not of the axis. In fact, ignore the axis: they don't exist, only their directions have mathematical meaning (neither their scale nor their points of origin mean anything, because the affine transformations of the utility functions will shift those).
Back in the old days, when people were wise and the government was just, I did a post on the Nash bargaining solution for two player games. Here each player has their own utility function and they're choosing amongst joint options, and trying to bargain to find the best one. What was nice about this solution is that it is independent of irrelevant alternatives (IIA): once you've found the best solution, you can erase any other option, and it remains the best.
In order to do that, the Nash bargaining solution makes use of a "disagreement point", a special point that provides a zero to both utilities. This seems - and is - ugly. Can we preserve IIA without this clunky disagreement point?
By the title of the this post, you may have guessed that we can't. Specifically, assume the outcome is symmetric across both players (i.e. permuting the two utility functions preserves the outcome choice), the outcome is Pareto-optimal (any change will reduce the utility of at least one player) and there is no outside canonical choices for the utility functions (no special scales, no zeroes, no disagreement points). Then IIA must fail. It fails under weaker conditions as well, but the above lead to an easy picture-proof. And picture proofs are nice.
So assume there are five possible choices, whose utility values for the two players are (0, 3), (1.2, 2.6), (2, 2), (2.6, 1.2), (3, 0). These are graphed here:
The choice set is symmetric and the green point (2, 2) is Pareto-optimal and on the axis of symmetry. Hence by the assumptions, the green point must be the outcome chosen. Now further assume IIA, and we will derive a contradiction.
First, by IIA, we can erase the losing points (2.6, 1.2) and (3, 0). Then we can rescale the utility functions: the utility function graphed on the x axis is divided by two, while the utility graphed on the y axis has 2 subtracted from it. These changes are illustrated here:
This results in a final setup of (0, 1), (0.6, 0.6) and (1, 0):
But this is obviously wrong: symmetry implies the correct outcome should be the blue point (0.6, 0.6), not the green (1, 0) which was the outcome before we removed the "irrelevant" extra points. We have derived a contradiction, and IIA must fall.