All definitions used in this post are from this paper. This post is inspired by the beginning of Section 2.2 in the same paper, and the following set-up for my question is cited from Section 2.2 in the same paper.

Let $G$ be a countable group that defines a non-singular action on a measure space $(X, \mu)$. Define the following mapping:

$$ \alpha: G\times X\rightarrow \mathbb{R}, \hspace{0.2cm} (g, x) \mapsto \log\frac{d\,g\mu}{d\,\mu}(x) = \frac{d\,\mu(g\,\cdot\,x)}{d\,\mu(x)} $$

The mapping $\alpha$ satisfies the $1$-cocycle identity:

$$ \alpha(gh, x) = \alpha(g, h\,\cdot\,x)+\alpha(h, x) $$

for all $g, h\in G$ and for $\mu$-almost every $x\in X$. Then for each $g\in G$, the group action defined on $X\times \mathbb{R}$:

$$ g\,\cdot\,(x, t) = (g\,\cdot\,x, \alpha(g, x)+t) $$

is called the Maharam extension of $G\curvearrowright(X, \mu)$. Observe that the measure $d\mu \times e^{-t}\,dt$ on $X\times \mathbb{R}$ is $G$-invariant. The $G$-action on $X\times \mathbb{R}$ commutes with the $\mathbb{R}$-action on $X\times \mathbb{R}$ where the latter one is defined by $s\,\cdot\,(x, t) = (x, s+t)$. Let $(Z, \eta)$ denote the space of ergodic components of $G\curvearrowright (X, \mu)$, so $L^{\infty} \big(X, \mu \big)^G$ the algebra of $G$-invariant function in $\big(X\times \mathbb{R}, d\mu \times e^{-t}dt\big)$ can be identified by $L^{\infty}\big(Z, \eta\big)$.

My question is the following: fix $\tau: G\times X\rightarrow \mathbb{R}$ another mapping that is different from $\alpha$ and satisfies the $1$-cocycle identity, and defines the following new $G$-action on $X\times \mathbb{R}$:

$$ g\,\ast\,(x, t) = (g\,\cdot\,x, \tau(g, x)+t ) $$ Fix an arbitrary $F\in L^{\infty}(X, \mu)^G$, so by definition we have $g\,\cdot\,F = F$ for any $g\in G$. If we assume that $g\,\ast\,F = F$ for all $g\in G$, what can we tell between the action $g\,\cdot\,(x,t)$ and $g\,\ast\,(x, t)$? Could they be the same, conjugate with each other or it all depends on the choice of $\tau$?