G$ is a union of disjoint cycles. It follows that $\mathcal G$ has a perfect matching, i.e., a set of $d$ edges which do not have a vertex in common. Using this matching we define $\tau\in S_d$ so that the edges of the matching are $(\mathbf{a}_s,\mathbf b_{\tau(s)})$. The label on each edge $(\mathbf{a}_s,\mathbf b_{\tau(s)})$ of the matching is the corresponding $b_{\tau(s)}$, and $a_s:=\sigma^{-1}\big(b_{\tau(s)}\big)$.

Now let $\sigma_1,\sigma_2\in H_{2d}$ be defined by $$\sigma_1(t):=\begin{cases}
2s-1& \text{ if }t=2\tau(s)-1\text{ for }1\leq s\leq d,\\
2s& \text{ if }t=2\tau(s)\text{ for }1\leq s\leq d.
\end{cases}
\ \ \text{ and }\ \
\sigma_2:=\prod_
{\substack{1\leq s\leq d\\[.3mm]
a_s-b_{\tau(s)}\not\in 2\Z}}\!\!\!
(2s-1,2s),$$ and set $\sigma_3:=\sigma_2\sigma_1\sigma$. It is easy to check that that $\sigma_3(a_s)=a_s$ for $1\leq s\leq d$. Next set $$\sigma_4:=\prod_{\substack{1\leq s\leq d\\ a_s=2s}}(2s-1,2s)\text{\ \,and\,\ }\sigma_5:=\sigma_4\sigma_3\sigma_4^{-1},$$ so that $\sigma_5(2s-1)=2s-1$ for $1\leq s\leq d$. Now let $\tau\in S_d$ be defined by $\tau(s):=\frac{1}{2}\sigma_5(2s)$ for $1\leq s\leq d$. Choose $0=d_0<d_1<\cdots<d_r=d$ and $\tau_\circ\in S_d$ such that $$\tau_\circ^{}\tau\tau_\circ^{-1}=
\prod_{s=1}^d(d_{s-1}+1,\ldots,d_s).$$ Finally, let $\sigma_\circ\in H_{2d}$ be defined by $$\sigma_\circ(2s):=
2\tau_\circ(s)
\,\text{ and }\,
\sigma_\circ(2s-1):=
2\tau_\circ(s)-1
\,\text{ for }\,1\leq s\leq d
.$$ It is straightforward to check that $\sigma_\circ^{}\sigma_5\sigma_\circ^{-1}
=\tau_{d_0,d_1}^{}\cdots \tau_{d_{r-1},d_r^{}}^{}$, so that holds for $\sigma':=\sigma_\circ^{}\sigma_4^{}\sigma_2^{}\sigma_1^{}$ and $\sigma'':=\sigma_4^{-1}\sigma_\circ^{-1}$.

\[prp-g-inv-sPWW\] For every $d\geq 1$, $$\label{sSdttsPd}
\left(
\sP^d(W)\otimes \sS^d(W)\right)^\g g 
=\mathrm{Span}_\C\left\{\bft_\sigma\,:\,\sigma\in S_{2d}\right\}
