Suppose $\circ:G \times X \to X$ with $X$ countable. There is a measure preserving dynamical system $([0,1]^X, \lambda^X, G)$ where $\lambda$ is the uniform measure on $[0,1]$ and for $(a\_x)\_{x \in X} \in [0,1]^X$, $g$ takes $(a\_x)\_{x \in X}$ to $(a\_{\circ(g,x)})\_{x \in X}$. 


If $X$ has a finite orbit $Y$ then there is an invariant measure preserving map from $([0,1]^X, \lambda^X)$ to $([0,1], \lambda)$. For example


$$(a\_x)\_{x \in X} \mapsto \sum\_{y \in Y} a\_y \mod 1.$$


If $X$ has no finite orbits will there still always be an invariant measure preserving map from $([0,1]^X, \lambda^X)$ to $([0,1], \lambda)$?


As a concrete example, is there such an invariant measure preserving map when $X = (\mathbb{Z}, S)$ and $G = $Aut$(\mathbb{Z}, S)$, where $S$ is the successor relation (and there is no other structure on $\mathbb{Z}$)?



Is there always an G-invariant uniform measure preserving map for countable G?
I presume that by an "invariant" map you mean one which is constant on the orbits of the action of $G$ on the configuration space. Therefore, it should factorize through the space of ergodic components of your action, and your question is equivalent to asking whether the space of ergodic components is purely non-atomic. 


In the case when $X=\mathbb Z$ and $G=Aut(\mathbb Z)$ already the action of $\mathbb Z$ is ergodic (essentially by Kolmogorov's 0-1 law), so that the answer is "no".


