Suppose, the group $ G=S(2^{\infty})$ has a unitary representation $ \pi $ on a separable infinite dimensional Hilbert space $ H $.
(The group $ S(2^{\infty}) $ is the direct limit of the following directed system of symmetric groups. $ S_1\to S_2\to S_4\to .....\to S_{2^n}\to S_{2^{n+1}}\to.... $ where the map from $ S_{2^i} $ to $ S_{2^{i+1}} $ is given by the composition of the maps $ S_{2^i}\overset{x\to (x,x)}{\longrightarrow} S_{2^i}\times S_{2^i} $ and $ S_{2^i}\times S_{2^i}\hookrightarrow S_{2^{i+1}} $)
If $ G $ has a unitary representation on $ H $, this gives unitary representation of $ S_{2^n} $ on $ H $ for each $ n $(through the inclusion $ S_{2^n}\hookrightarrow G $ coming from the directed system in the definition of $ S(2^{\infty})$).
Suppose, for every $n $, $ H $ is isomorphic to $ \mathbb{C}[S_{2^n}]\otimes L$ as $ S_{2^n} $ module where $ \mathbb{C}[S_{2^n}] $ is the regular representation of $ S_{2^n} $ and $ L $ is a infinite dimensional separable Hilbert space with $ S_{2^n} $ acting trivially on it. Consider, the representation $ \pi\otimes \pi $ of $ G $ on $ H\otimes H $.
Then, is there a $ G $ module isometry from $ H $ to $ H\otimes H $, i.e., an isometry $ S: H\to H\otimes H $ so that $ (\pi(g)\otimes \pi(g))S=S\pi(g) $ for every $ g\in G $.
