The answer to [this question][1] implies that each locally compact group topology $\tau$ on the permutation group $S^\infty$ is discrete. Since the discrete group $S^\infty$ is known to be non-amenable (it contains a free grouo with two generators), the locally compact topological group $(S^\infty,\tau)$ is not amenable.

By **Corollary** in the answer to [this question][1], the permutation group $S^\infty$ is not isomorphic to a dense subgroup of a non-discrete locally compact group. This implies that $S^\infty$ is not isomorphic to a dense subgroup of an amenable locally compact group.

  [1]: https://mathoverflow.net/questions/293614/is-each-locally-compact-group-topology-on-the-permutation-group-discrete/293619#293619