Let $g(x,0) = x$ and $g(x,t+1) = g(x,t) - \dfrac{1}{g(x,t)}$ for every real $t$.
From the fact
\begin{align} \int_{-\infty}^{\infty}f\left(x-\frac{1}{x}\right)dx&=\int_{0}^{\infty}f\left(x-\frac{1}{x}\right)dx+\int_{-\infty}^{0}f\left(x-\frac{1}{x}\right)dx=\\ &=\int_{-\infty}^{\infty}f(2\sinh T)\,e^{T}dT + \int_{-\infty}^{\infty}f(2\sinh T)\,e^{-T}dT=\\ (collecting\space terms ) &=\int_{-\infty}^{\infty}f(2\sinh T)\,2\cosh T\,d T=\\ &=\int_{-\infty}^{\infty}f(x)\,dx. \end{align}
as discussed here : Why is this integral $\int_{-\infty}^{+\infty} F(f(x)) - F(x) dx = 0$?
I am tempted to conclude
$\int_{-\infty}^{\infty}f(g(x,t))=\int_{-\infty}^{\infty}f(x)\,dx.$
for every real $t$.
SO I wondered about the general situation :
Let $H(x)$ be some real-analytic function such that :
$\int_{-\infty}^{\infty}f(H(x))=\int_{-\infty}^{\infty}f(x)\,dx.$
and the integral exists,
and define,
$G(x,0) = x$ and $G(x,t+1) = H(G(x,t)) $ for every real $t$.
Does this Always imply that :
$\int_{-\infty}^{\infty}f(G(x,t))=\int_{-\infty}^{\infty}f(x)\,dx.$ ?
Related :
$\int_0^{\infty} A( f(B(x)) ) + C(x) ) dx = \int_0^{\infty} f(x) dx$
Why is this integral $\int_{-\infty}^{+\infty} F(f(x)) - F(x) dx = 0$?
$$edit :$$
I avoided talking about uniqueness to avoid making things to complicated , but when uniqueness of the inverse Abel functions seems an issue then the truth of this conjecture might give an intresting uniqueness criterion for complex dynamics ?
Btw notice $x-\frac{1}{x}$ has its fixpoint at infinity.
I assume not all of the real fixpoints of $H(x)$ can be parabolic. Also the inverse Abel function for $H$ will not use real parabolic fixpoints as the correct way of computation.
IN FACT probably $H(x)$ never has real fixpoints to avoid the PROBLEMATIC case of convergeance : $f(x) , f(H(x)) , f(H(H(x))) , ... $ $f$(fixpoint) = nonzero constant !! Although that is not a proof.
I must give some credit to my mentor tommy1729 who talked about this last Friday.
$$EDIT\space 2 :$$
Too adress sheldon's comment :
To compute $g(x,\frac{1}{2})$ I use the following method :
$g(x,\frac{1}{2}) = \lim_{n\space \to +\infty} g(\frac{g(x,-n)+g(x,-n+1)}{2},n) $
More generally
$g(x,t) = \lim_{n\space \to +\infty} g((1-t)\space g(x,-n)+ t\space g(x,-n+1),n) $
$$EDIT \space 3$$
It might be intresting to note that a neccessary condition for the conjecture to be true seems to be this below
$$\frac{d}{dt} \space \int_{-\infty}^{\infty}f(g(x,t))= 0$$
One can then use differentiation under the integral sign and the chain rule but it seems like an alternative statement rather then a step closer to a solution ?
Im still playing with this idea.