Let $X$ be a compact Hausdorff space where $X$ have infinitely many points and the topology is non-discrete, $m$ be a regular probability measure defined on the Borel $\sigma$-algebra of $X$, and $g$ an homeomorphism of $X$. Suppose that the image measure $g\_{\ast}m$ (defined by $g\_{\ast}m(O) = m\big[g^{-1}(O)\big]$ for each Borel subset $O$) is equivalent to $m$. Then, for each $F\in C(X)$, I will have:


$$
\begin{aligned}
&\hspace{0.6cm} \int\_X F(x)\frac{d\,g\_{\ast}m}{d\,m}(x)\,dm(x) = \int\_X F(x)\,d(g\_{\ast}m)(x)\\
& = \int\_X F(x)\, dm(g^{-1}(x)) = \int\_X F\circ g(x)\,dm(x)
\end{aligned}
$$


Define $\phi\_m$ to be the positive linear functional defined by $m$ (and defined on $C(X)$) and $\phi\_{g\_{\ast}m}$ similarly, $I = \{F\in C(X): \phi\_m(\vert\,f\,\vert)=0\}$ and $I\_g = \{F\in C(X): \phi\_{g\_{\ast}m}(\vert\,f\,\vert) = 0\}$. Then the image of the following mapping:


$$
M: I\_g\rightarrow I,\hspace{0.3cm} F(x) \mapsto F(x)\frac{d\,g\_{\ast}m}{d\,m}(x)
$$
is equal to the image of the action (restrcited on $I\_g$) defined by $g\bullet F(x) = F\circ g(x)$. Obviously both mappings are invertible and the action is a isometric isomorphism between $I\_g$ and $I$. My questions are:


1. Since the action $F\mapsto g\bullet F$ is multiplicative, $M$ will never coincide with the action unless $g$ is the identity. In this case what can we tell about the local range of $(d\,g\_{\ast}m/d\,m)$?
2. Is it true that, when $X$ has non-empty homeomorphism group, for each regular probability measure $m$, there exists a non-identity element $g\in\operatorname{Homeo}(X)$ (the homeomorphism group of $X$) such that $m$ is equivalent to $g\_{\ast}m$?
3. Given a non-trivial countable subgroup $G\subseteq\operatorname{Homeo}(X)$, will there exist a regular non-atomic probability measure $m$ such that $g\_{\ast}m\sim m$ for each $g\in G$?


**Update**: Let $m$ be a probability regular non-atomic measure and assume $g\in\operatorname{Homeo}(X)$ satisfies $m\sim gm$. According to this [post](https://math.stackexchange.com/questions/417723), we can find a compact set $C\_{\epsilon}$ such that both $m(C\_{\epsilon})$ and $gm(C\_{\epsilon})$ are less than $\epsilon$. I tried to use this and the **Urysohn's Lemma** to decide the local behavior of the derivative but failed. Any hints will be appreciated.


**Update2**: From the **Corollary 6.2.2** in **Measures on topological space** written by *V.I.Bogachev* ([link](https://link.springer.com/article/10.1007/bf02432851)), for a fixed $g\in\operatorname{Homeo}(X)$, one can find a Radon probability measure $m$ (and hence regular when $X$ is compact) such that $g\_{\ast}m = m$. In such case, compared to my original question, I suppose it would not be too greedy to ask, given $m$ a probability measure, is it possible to find a subgroup $G\subseteq\operatorname{Homeo}(X)$ such that $g\_{\ast}m\sim m$ for each $g\in G$.



Radon-Nikodym derivative in a compact Hausdorff space
I think the answer for 2 is **no**. I will state some counter-examples below, after some partial results.


A Necessary condition involving the support
-------------------------------------------


From now on, all measures are Borel probability measures. For every measure $\mu$, define its support
$$\mathrm{Supp}(\mu) := \overline{\{ x : \forall V \text{ open with } x \in V, \mu(V)>0 \}} .$$


We have $\mathrm{Supp}(g\_\*\mu) = g(\mathrm{Supp}(\mu))$.
Indeed, let $x$ be in $\mathrm{Supp}(g\_\*\mu)$, then for every $V \ni x$ open, $0<g\_\*\mu(V) = \mu(g^{-1}(V))$. From there we can see that $g^{-1}(x) \in \mathrm{Supp}(\mu)$.


An immediate consequence of the definition is that two equivalent measures have the same support. This gives us the following Lemma:


**Lemma:** *For every homeomorphism $g$, if $\mu$ and $g\_\*\mu$ are equivalent, then $g(\mathrm{Supp}(\mu)) = \mathrm{Supp}(\mu)$.*


Partial answer for measures supported on a strict subspace
----------------------------------------------------------


**Proposition 1 (positive answer to 2.):** *If for every open set $V \notin \{\emptyset, X\}$, there is a homeomorphism $g$ of $X$ such that $g(V) = V$ and $g|\_V \neq id$, then for every $\mu$ with $ \mathrm{Supp}(\mu) \neq X$ we have a non-identity $g$ such that $\mu$ and $g\_\*\mu$ are equivalent.*


Indeed, fix $V = X \setminus \mathrm{Supp}(\mu)$, and take $g$ given by the assumption. The assumption of Proposition 1 is true on many obvious examples.
Note the condition that $\mathrm{Supp}(\mu)\neq X$. See the conjecture below that relates to the case where we do not have this condition. The next proposition provides a partial converse to Proposition 1.


**Proposition 2 (negative answer to 2.):** *Assume that $X$ is separable. If there exists an open set $V \notin \{\emptyset, X\}$ such that the only homeomorphism $g$ with $g(V)=V$ is the identity, then there is a measure $\mu$ such that the only homeomorphism $h$ with such that $\mu$ and $h\_\*\mu$ are equivalent is the identity.*


We use the separability of $X$ (more precisely of $V^c$) to explicitely construct $\mu$. Maybe we can still construct $\mu$ under weaker conditions, or conversely find some non-separable $X$ for which Proposition 2 fails. A look at [examples of non-separable compact spaces](https://math.stackexchange.com/questions/785087/non-separable-compact-space) (which are necessarily not metrizable) might help.


*proof:* The challenge consists in finding a suitable $\mu$. Since $X$ is separable, so is the compact space $V^c$: let $(v\_n)\_{n\geq 1}$ be sequence of $V^c$ that is dense in $V^c$, and $\mu = \sum\_{n\geq 1} 2^{-n} \delta\_{v\_n}$. We can check that $\mu$ is a regular Borel measure with support $V^c$. Any homeomorphism $g$ such that $\mu$ and $g\_\*\mu$ are equivalent must have $V^c = g(V^c) \iff V = g(V)$, thus are the identity. $\square$


Examples of $X$ satisfying the assumption of Proposition 2 include spaces with an isolated point. Here is a somewhat complicated example without isolated point. We fix a numbering $(q\_n)\_{n\geq 1}$ of $\mathbb Q \cap (0,1)$, and construct a space $T$ as the limit of the following inductive construction --- which should remind you of Koch’s snowflake:


* start with $T = [0,1]$, and "distinguish" the sequence $(q\_n)$.
* For every $n\geq 1$, glue $2n$ copies of $[0,2^{-n}]$ (with distinguished vertices $(2^{-n} q\_k)\_{k\geq 1}$) to the point $q\_n$, by identifying the zeroes of the copies with $q\_n$. This way, the $q\_n$ all have non-homeomorphic neighborhoods ($q\_n$’s neighborhood looks like a $2n+2$-pointed star), so any homeomorphism of $T$ must send $q\_n$ to $q\_n$ for every $n$. This ensures that the homeomorphisms of $T$ are the identity on the original copy of $[0,1]$.
* Iterate this process on the glued copies, making sure that the added copies are small enough so that everything converges **and that no distinguished vertex ever has the same number of branches than some other distinguished vertex.**


You should obtain $T$ that is a compact tree with infinitely many branches, with a countable number of branching points (the distinguished vertices) that are dense in $T$, such that the only homeomorphism of $T$ is the identity.


Now build $X$ by taking two copies $T\_1$ and $T\_2$ of $T$ and identifying the zero of $T\_1$ with the one of $T\_2$ and the one of $T\_1$ with the zero of $T\_2$. There are two homeomorphisms of $X$: the identity, and the exchange sending $T\_1$ onto $T\_2$. But once we fix any measure with support included in $T\_1$, the only homeomorphism $g$ that ensures that $\mu$ and $g\_\*\mu$ are equivalent is the identity.


Conjectured negative example with full support
----------------------------------------------


Finally, here is a conjecture that, if true, provides a "natural" X and a rather natural $\mu$ with support $X$ for which 2. fails.


**Conjecture:** *Let $X=[0,1]$ equipped with the usual topology. The only homeomorphism that leaves $K = (\mathbb Q \cap [0,1]) \cup \{e^q, q\in \mathbb Q \cap (-\infty, 0]\}$ is the identity.*


Note that $X$ has a non-trivial homeomorphism group. $K$ is countable; let $(k\_n)\_{n\geq 1}$ be an enumeration of $K$, and define $\mu = \sum\_{n\geq 1} 2^{-n} \delta\_{k\_n}$ --- we can check that it is Borel and regular. For every homeomorphism $g$ of $X$, $\mu$ and $g\_\*\mu$ are equivalent if and only if $g(K) = K$.
Then assuming this conjecture, the only homeomorphism with $\mu$ and $g\_\*\mu$ equivalent is the identity, providing a negative answer to 2. even for a "nice" $X$.


