It's easy to see that the action is always ergodic since $\mathcal G$ contains the group of finite permutations on the indices, which acts ergodically. In fact, the group $\mathcal G$ ($= \mathcal which equals$\mathcal H$in the case$m = \mu^{\mathbb N}$with$\mu(0) = 1/2$.) that you are describing is the full group of the ergodic hyperfinite measurable equivalence relation. It, and other full groups are discussed in Sections I.3 and I.4 in the book by Alexander Kechris: Global aspects of ergodic group actions, Mathematical Surveys and Monographs, 160, American Mathematical Society, 2010. 1 It's easy to see that the action is always ergodic since$\mathcal G$contains the group of finite permutations on the indices, which acts ergodically. In fact, the group$\mathcal G$($= \mathcal H\$) that you are describing is the full group of the ergodic hyperfinite measurable equivalence relation. It, and other full groups are discussed in Sections I.3 and I.4 in the book by Alexander Kechris: Global aspects of ergodic group actions, Mathematical Surveys and Monographs, 160, American Mathematical Society, 2010.