MATH.9:5.4 - Choose a position despite an ambiguous normalization

Let the six permutations of positions 1, 2 and 3 act on the triples formed by permuting d0=(a,a,b), where a and b differ. An answer may be any of the three positions, and relabeling the triple must relabel the answer.

The transformations fixing d0 are the identity and the swap of positions 1 and 2. All three positions are permitted answers, but only position 3 is fixed by both transformations. Thus B(d0)={3}; choose y0=3.

For the input d=(a,b,a), two transformations normalize it to d0. The first, h1, swaps positions 2 and 3. The second, h2, sends positions 1 to 2, 2 to 3 and 3 to 1. Both carry the b entry to position 3 while placing the two a entries in the remaining positions.

Returning y0 uses their inverses:

h1^-1(3)=2=h2^-1(3).

For every permutation of d0, the transported answer is the position carrying b. This gives the rule on the entire three-input orbit. The stabilizer calculation makes the return independent of the normalizer even though the input initially permits three different answers.