Let $X$ be a compact Hausdorff space where $X$ have infinitely many points and the topology is non-discrete, $m$ be a regular non-atomic probability measure defined on the Borel $\sigma$-algebra of $X$, and $g$ an autohomeomorphism 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:

$$ \int_X F(x)\frac{d\,g_{\ast}m}{d\,m}(x)\,dm(x) = \int_X F(x)\,d(g_{\ast}m)(x) = \int_X F(x)\, dm(g^{-1}(x)) = \int_X F\circ g(x)\,dm(x) $$

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, for each probability regular non-atomic 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, 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), 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 $g\in G$.

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, 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), 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 $g\in G$.

Let $X$ be a compact Hausdorff space, $m$ be a regular non-atomic probability measure defined on the Borel $\sigma$-algebra of $X$, and $g$ an autohomeomorphism 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:

$$ \int_X F(x)\frac{d\,g_{\ast}m}{d\,m}(x)\,dm(x) = \int_X F(x)\,d(g_{\ast}m)(x) = \int_X F(x)\, dm(g^{-1}(x)) = \int_X F\circ g(x)\,dm(x) $$

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, for each probability regular non-atomic 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 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, 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), 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 $g\in G$.

Let $X$ be a compact Hausdorff space where $X$ have infinitely many points and the topology is non-discrete, $m$ be a regular non-atomic probability measure defined on the Borel $\sigma$-algebra of $X$, and $g$ an autohomeomorphism 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:

$$ \int_X F(x)\frac{d\,g_{\ast}m}{d\,m}(x)\,dm(x) = \int_X F(x)\,d(g_{\ast}m)(x) = \int_X F(x)\, dm(g^{-1}(x)) = \int_X F\circ g(x)\,dm(x) $$

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, for each probability regular non-atomic 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, 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), 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 $g\in G$.