Let $\mathcal{B}$ be the (unique up to isomorphism) countable atomless boolean algebra, and $\mathrm{Aut}(\mathcal{B})$ its automorphism group, with pointwise convergence topology.

My question: Does $\mathrm{Aut}(\mathcal{B})$ contain a **closed** (non-abelian) free subgroup?

This question arises from reading about the following two results (whose proofs I am not overly familiar with):

$\mathrm{Aut}(\mathcal{B})$ contains many free subgroups; by a result of Machpherson (*Groups of Automorphisms of $\aleph_0$-Categorical Structures*, QJM, 1986), it has a dense subgroup freely generated by countably many generators.

$\mathrm{Aut}(\mathbb{Q},<)$ contains a closed non-abelian free group; this is a result of Pestov (*On free actions, minimal flows, and a problem by Ellis*, TAMS, 1998). In particular, closed subgroups of extremely amenable topological groups need not be amenable.