Let $X=\{0,1\}^{\mathbb{N}}$ and $\xi\_n$ be the partition of $X$ defined by the equivalence relation $x \sim\_n x' \Leftrightarrow (x\_{n}, x\_{n+1}, \ldots) = (x\_{n}', x\_{n+1}', \ldots)$. The sequence of partitions $(\xi\_n)$ is decreasing and we introduce the intersection partition $\theta= \cap \xi\_n$. I'm loooking for a group $G$ of transformations of $X$ such that $\theta$ is the partition into the orbits of $G$. 


Maybe my question is somewhat unprecise, I am rather new in ergodic theory. Any comments are welcomed.



intersection partition as an orbital partition
Consider the group $G\_n$ of all permutations of strings $\{0,1\}^{\{1,2,...,n\}}$ of length $n$. That group acts on the set of all strings $\{0,1\}^{\mathbb N}$ in the natural way (it changes the $n$-prefix only). Now clearly $G\_n\subset G\_{n+1}$ for every $n$. Let $G=\cup G\_n$. It is a (locally finite) group which acts on $\{0,1\}^{\mathbb{N}}$ and the orbit equivalence relation is your $\theta$ (if I understood the question correctly): two strings are in the same orbit iff they coincide almost everywhere. 


