Gallavotti-Cohen action functional

Question

For a time continuous Markov jump process a path $\left\{\sigma_{s}, 0 \leqslant s \leqslant t\right\}$ is time reversed as $\left\{\sigma_{t-s}, 0 \leqslant s \leqslant t\right\}$. The time reversed process has the jump rates $k^{R}\left(\sigma, \sigma^{\prime}\right)=k\left(\sigma^{\prime}, \sigma\right) \mu_{s}\left(\sigma^{\prime}\right) / \mu_{s}(\sigma)$ whereas the inverse waiting times, $r(\sigma)=\sum_{\sigma'}k(\sigma,\sigma')$, and the steady state, $\mu_{s}(\sigma)$, remain unmodified. If $P_{[0, t]}$ denotes the path measure of the stationary process in the time window $[0, t]$ and $P_{[0, t]}^{R}$ the one of the corresponding time reversed process, then $P_{[0, t]}^{R}$ has a density relative to $P_{[0, t]}$ and \begin{equation}\label{1} P_{[0, t]}^{R}=e^{-W(t)}\left(\mu_{s}\left(\sigma_{t}\right) / \mu_{s}\left(\sigma_{0}\right)\right) P_{[0, t]},\quad\quad(1) \end{equation} where $W(t)$ is the action functional. This functional is defined in the following way: $$W\left(t,\left\{\sigma_{s}, 0 \leqslant s \leqslant t\right\}\right)=\int_{0}^{t} \sum_{\sigma, \sigma^{\prime}} w_{\sigma, \sigma^{\prime}}(s) d s,$$ with $w_{\sigma, \sigma^{\prime}}(s)$ a sequence of $\delta$ -functions, located exactly at those times $s$ when $\sigma_{s}$ jumps from $\sigma$ to $\sigma^{\prime}$, with weight $$ w\left(\sigma, \sigma^{\prime}\right)=\log k\left(\sigma, \sigma^{\prime}\right)-\log k\left(\sigma^{\prime}, \sigma\right). $$ We have therefore that, up to boundary terms, the action functional equals $-\log \left(d P_{[[0, t]}^{R} / d P_{[0, t]}\right)$ with $d P^{R} / d P$ denoting the Radon-Nikodym derivative. Now, we want to see what happens if instead of the time-reversal transformation we use another transformation.If we consider an internal symmetry $S: \mathscr{S} \rightarrow \mathscr{S}$ such that $S \circ S=1$ and we use this transformation in $(1)$, then the GC action functional becomes $$ W(t)=-\int_{0}^{t} d s \log \left[r\left(S \sigma_{s}\right) / r\left(\sigma_{s}\right)\right]\quad\quad (2) $$ up to boundary terms. Why $(2)$ is true?