Write a fair permutation in cycle form, as follows. First write all cycles of length $>1$ in decreasing order of their smallest element, with the smallest element of each cycle written as the leftmost element of the cycle. Then append all the fixed points in increasing order. An example is $$ (3,9,4,6)(1,7,2,10)(5)(8). $$ Now erase the parentheses. We obtain a Salié permutation. Each Salié permutation arises in exactly two ways. If the first fixed point is less than the element preceding it in the representation just described (as is the case for the example above), then we can absorb the first two fixed points into the cycle preceding them (that is, the rightmost cycle). Otherwise, we can turn the last two elements of the rightmost cycle into fixed points. For the example above we get the additional fair permutation $$ (3,9,4,6)(1,7,2,10,5,8) $$ yielding the same Sailé permutation.

Write a fair permutation in cycle form, as follows. First write all cycles of length $>1$ in decreasing order of their smallest element, with the smallest element of each cycle written as the leftmost element of the cycle. Then append all the fixed points in increasing order. An example is $$ (3,9,4,6)(1,7,2,10)(5)(8). $$ Now erase the parentheses. We obtain a Salié permutation. Each Salié permutation arises in exactly two ways. If the first fixed point is less than the element preceding it in the representation just described (as is the case for the example above), then we can absorb the first two fixed points into the cycle preceding them (that is, the rightmost cycle). Otherwise, we can turn the last two elements of the rightmost cycle into fixed points. For the example above we get the additional fair permutation $$ (3,9,4,6)(1,7,2,10,5,8) $$ yielding the same Sailé permutation.

Addendum. There is a small inaccuracy above. If 1 is a fixed point, then rather than absorbing 1 and the fixed point $j$ following it into the previous cycle, we should create a new cycle $(1,j)$.

Write a fair permutation in cycle form, as follows. First write all cycles of length $>1$ in decreasing order of their smallest element, with the smallest element of each cycle written as the leftmost element of the cycle. Then append all the fixed points in increasing order. An example is $$ (3,9,4,6)(1,7,2,10)(5)(8). $$ Now erase the parentheses. We obtain a Salié permutation. Each Salie permutation arises in exactly two ways. If the first fixed point is less than the element preceding it in the representation just described (as is the case for the example above), then we can absorb the first two fixed point into the cycle preceding them (that is, the rightmost cycle). Otherwise, we can turn the last two elements of the rightmost cycle into fixed points. For the example above we get the additional fair permutation $$ (3,9,4,6)(1,7,2,10,5,8) $$ yielding the same Sailé permutation.

Write a fair permutation in cycle form, as follows. First write all cycles of length $>1$ in decreasing order of their smallest element, with the smallest element of each cycle written as the leftmost element of the cycle. Then append all the fixed points in increasing order. An example is $$ (3,9,4,6)(1,7,2,10)(5)(8). $$ Now erase the parentheses. We obtain a Salié permutation. Each Salié permutation arises in exactly two ways. If the first fixed point is less than the element preceding it in the representation just described (as is the case for the example above), then we can absorb the first two fixed points into the cycle preceding them (that is, the rightmost cycle). Otherwise, we can turn the last two elements of the rightmost cycle into fixed points. For the example above we get the additional fair permutation $$ (3,9,4,6)(1,7,2,10,5,8) $$ yielding the same Sailé permutation.