Free action of $\mathbb{Z}(2^{\infty})$ on a compact space

Question

Assume that $X$ is a Hausdorff compact space such that $\forall n\in \mathbb{N}$, we have a free action of $\mathbb{Z/{2^{n}}\mathbb{Z}}$ on $X$. Must $\mathbb{Z}(2^{\infty})$ act freely on $X$?

Note that we do not assume any compatibility of these actions with the inclusion $\mathbb{Z/{2^{n}}\mathbb{Z}}$ in $\mathbb{Z/{2^{n+1}}\mathbb{Z}}$. That is: the action of $\mathbb{Z/{2^{n}}\mathbb{Z}}$ on $X$ is not necessarily the same action of $\mathbb{Z/{2^{n}}\mathbb{Z}}$ as a subgroup of $\mathbb{Z/{2^{n+1}}\mathbb{Z}}$.

Answer 1

I tried to find a counterexample, but ended up with the following interesting example instead. Put $X=\prod_{n=0}^\infty\{0,1\}$. Define $$ \phi_n : X \to \mathbb{Z}/2^n \times X $$ by $$ \phi_n(a) = \left(\sum_{i=0}^{n-1}2^{n-1-i}a_i,(a_n,a_{n+1},\dotsc) \right). $$ Define $\tau_n:\mathbb{Z}/2^n\to\mathbb{Z}/2^n$ by $\tau_n(i)=i+1$, then define $\sigma_n:X\to X$ by $\sigma_n=\phi_n^{-1}\circ(\tau_n\times 1)\circ\phi_n$. One can check that $\sigma_n^{2^n}=1$ and $\sigma_{n+1}^2=\sigma_n$, so the maps $\sigma_n$ fit together to give an action of $\mathbb{Z}/2^\infty$ on $X$. This is easily seen to be free.