Are all Bernoulli measure preserving homeomorphisms of the Cantor space eventually tree automorphisms?

Question

Setting: Consider the Cantor space $X=\{0,1\}^\mathbb{N}$. For $n\in\mathbb{N}$ and $\bar{x}\in\{0,1\}^n$, we define the cylinder sets $Z_{\bar{x}}:=\{x\in X: (x_1,\dots,x_n)=\bar{x}\}$, sot that $X=\bigsqcup_{\bar{x}\in\{0,1\}^n}Z_{\bar{x}}$.

Let's say that a homeomorphism $h:X\to X$ is eventually a tree automorphism when $h$ has the property that, there is $n_0\in\mathbb{N}$ such that, for any $n\ge n_0$ and any $\bar{x}\in\{0,1\}^n$, there is a (necessarily unique) $\bar{y}\in\{0,1\}^n$ such that $h(Z_{\bar{x}})=Z_{\bar{y}}$.

In other words, if one draws the standard picture of the Cantor space as the leaves of the rooted binary tree (the nodes of which are the cylinder sets), $h$ is eventually a tree automorphism when from some level and on, $h$ induces a permutation of the nodes at each of the node levels.

Consider also the Borel probability measure $\mu$ on $X$ that is the uniform Bernoulli measure, i.e. $\mu$ is the infinite product of the normalized counting measure on $\{0,1\}$, and note that $\mu(Z_{\bar{x}})=2^{-n}$ for any $\bar{x}\in\{0,1\}^n$.

Question: Is it true that any homeomorphism $h\in\mathrm{homeo}(X)$ that preserves $\mu$ (in the sense that $\mu(h(E))=\mu(E)$ for all borel $E \subset X$) is eventually a tree automorphism? If not, can we explicitly write down a homeomorphism preserving $\mu$ that is not eventually a tree automorphism?

Remarks: Note that any homeomorphism $h\in\mathrm{homeo}(X)$ must preserve clopen sets, so the image of a cylinder set is necessarily a finite union of cylinder sets. If $h$ is also preserving $\mu$, the condition $\mu(Z_{\bar{x}})=2^{-n}$ creates more restrictions.

I've been drawing some binary trees (representing the semilattice of the cylinder sets) trying to come up with such a homeomorphism but I can't seem to spot one. Any ideas? Feel free to edit the tags.

Answer 1

No. Let $h((x_n)) = (y_n)$ where $y_{2n-1} = x_{2n}, y_{2n} = x_{2n-1}$.

Then $h$ is a homeomorphism of $X$ such that $\mu(h(E)) = \mu(E)$ for any Borel $E\subseteq X$.

Yet $h(Z_{\overline{x}})$ is not of the form $Z_{\overline{y}}$ for any odd $n$.

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More generally any bijection $f:\mathbb{N}\to\mathbb{N}$ will induce a homeomorphism $h_f((x_n)) = (x_{f(n)})$ such that $\mu(h_f(E)) = \mu(E)$ for any Borel $E$. This gives plenty of counter-examples.

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For as to why $\mu(h_f(E)) = \mu(E)$, recall this question, in particular this answer. The measure $\nu:E\mapsto \mu(h_f(E))$ is such that $$\nu(x_{i_1} = j_1, ..., x_{i_n} = j_n) = \mu(x_{f(i_1)} = j_1, ..., x_{f(i_n)} = j_n) = 2^{-n}$$ for all $i_1, ..., i_n\in\mathbb{N}$ and $j_1, ..., j_n\in\{0, 1\}$ so that $$\nu(x_{i_1} = j_1, ..., x_{i_n} = j_n) = \nu(x_{i_1} = j_1)...\nu(x_{i_n} = j_n)$$ so that the projections $\pi_i:X\to \{0, 1\}$ are independent random variables, and the pushforward with respect to $\pi_i$, that is the measure $k\mapsto \nu(x_i = k)$ is the measure on $\{0, 1\}$ which assigns probability $1/2$ to $0$, and $1/2$ to $1$. From uniqueness of products of probability measures, it follows that $\mu = \nu$ on the Borel $\sigma$-algebra.