3-bit Walsh permutation; conjugacy class 2+2

The elements in this conjugacy class are self-inverse and leave half of the elements fixed (i.e. there are two fixed points for one transposition).

The cardinality of this conjugacy class is determined by  A134057(n) = ${\binom {2^{n}-1}{2}}$. For n = 3 that is ${\binom {7}{2}}$ = 21, the number of 2-subsets of a 7-set.

The equations $a\oplus b~=~c$ show a bijection between the permutations and the 2-subsets of $\{1..7\}$.   ($\oplus$ is the bitwise XOR.)
$\{a,b,c\}$ are the positive fixed points.
$c$ is the XOR of the elements in any of the two cycles. The other two fixed points $\{a,b\}$ are one of the 21 subsets.
E.g. 134 has the cycles (2, 3) and (6, 7). 2⊕3 = 6⊕7 = 1, which is one of the fixed points. The other two are {4, 5}, so this is the subset.

The pattern of fixed and moved points corresponds to the rows 1..7 of a Walsh matrix. Permutations with the same fixed points are in the same column.

1	2	4	3	5	6	7
$4\oplus 6~=~2$	$1\oplus 5~=~4$	$2\oplus 3~=~1$	$4\oplus 7~=~3$	$2\oplus 7~=~5$	$1\oplus 7~=~6$	Walsh matrix 0 1 2 3 4 5 6 7 0 1 134 125 135 2 324 126 326 3 214 127 217 4 524 164 564 5 174 421 471 6 724 142 742 7 654 623 153     $a\oplus b~=~c$     $\{0,a,b,c\}$
Paris			London
$2\oplus 6~=~4$	$4\oplus 5~=~1$	$1\oplus 3~=~2$	$3\oplus 7~=~4$	$5\oplus 7~=~2$	$6\oplus 7~=~1$
Paris			Florence
$2\oplus 4~=~6$	$1\oplus 4~=~5$	$1\oplus 2~=~3$	$3\oplus 4~=~7$	$2\oplus 5~=~7$	$1\oplus 6~=~7$	$3\oplus 5~=~6$	$3\oplus 6~=~5$	$5\oplus 6~=~3$
Rome			Hamburg			Berlin

Above  A134057(n) has been expressed as a binomial, but it can also be expressed as a product $(2^{n-1}-1)\cdot (2^{n}-1)$.     (3 · 7 for n = 3)
That is the number of 0s in positive rows and columns of a binary Walsh matrix.
In the Walsh matrix shown above the positive rows correspond to $a\oplus b$, and the positive columns to the patterns of fixed points.

conversion of the formula
${\binom {2^{n}-1}{2}}~=~{\frac {(2^{n}-1)!}{2\cdot (2^{n}-3)!}}~=~{\frac {(2^{n}-2)\cdot (2^{n}-1)}{2}}~=~{\frac {2^{2n}-2^{n}-2^{n+1}+2}{2}}~=~2^{2n-1}-2^{n-1}-2^{n}+1~=~(2^{n-1}-1)\cdot (2^{n}-1)$