Relationship between the induced measure on an orbit and Hausdorff measure on the orbit Assumptions


*

*Let $G$ be a locally compact group that acts continuously on $\mathbb{R}^n$. 

*Let $\mu$ be a right Haar measure on $G$.

*Let $F:\mathbb{R}^n \to \mathbb{R}^m$ ($m < n$) be a Lipchitz function whose $m$-dimensional Jacobian $J_m F$ does not vanish on a compact set $A \subseteq  \mathbb{R^n}$. 

*Let $Gx = \{gx : g \in G \} \subseteq \mathbb{R}^n$ denote the orbit of a point $x \in \mathbb{R}^n$. 

*Assume that $F(Gx) = F(x)$ for all $x \in \mathbb{R^n}$ and $g \in G$. 

*Assume that $F(Gx)=F(Gy)$ implies $Gx=Gy$ for all $x,y \in G$

*Fix $x_0 \in \mathbb{R}^n$ and note that $Gx_0 = F^{-1}(F(x_0))$. 

*Let $H_{x_0}$ denote the restriction of $n-m$ dimensional Hausdorff measure to the set $Gx_0 = F^{-1}(F(x_0))$. 

*Let $\nu_{x_0}$ be the measure on $Gx_0 = F^{-1}(F(x_0))$ induced by $\mu$:
$$
\int_{Gx_0} f(x) d\nu_{x_0}(x) = \int_{G} f(gx_0) d\mu(g)
$$
Question: 
What is the relationship between $\nu_{x_0}$ and $H_{x_0}$? Are they absolutely continuous with respect to each other? If so, what properties does the density have? Are they constant multiples of each other?  
Edit/Update: 
user120527 has shown below that $G$ acting continuously on $\mathbb{R}^n$ is not enough to get $\nu_{x_0} \ll H_{x_0}$, and suggests additional regularity of the action as the fix. Since the action in the construction is asymmetric, I wonder whether the absolute continuity can be obtained by making the action symmetric.
In other words, what happens if we add the following assumption? 


*

*$G$ acts continuously on $\mathbb{R}$, and $G$ acts on $\mathbb{R}^n$ pointwise: $gx = (g x_1,\ldots,g x_n)$. 


I ask this as a separate question here: Take 2: Relationship between the induced measure on an orbit and Hausdorff measure on the orbit 
 A: Let's check an example.
Choose $\phi:\mathbb{R}\to \mathbb{R}$ be a increasing homeomorphism (to be chosen later).
Let $G=\mathbb{R}$ acting on $\mathbb{R}^2$ the following way:
$$t \cdot (x,y)= (\phi(\phi^{-1}(x)+t),y).$$
That is simply a conjugacy with the action by translation on the first coordinate. The thing is, thanks to the conjugacy by $\phi$, the points may move to the right in a weird, non-differentiable way.
The orbits $G(x_0,y_0)$ are simply the horizontal lines. Define $F(x,y)=y$. Then $H_{(x_0,y_0)}$ is simply the Lebesgue measure on the horizontal line $y=y_0$.
Now let us compute $\nu_{(x_0,y_0)}$ :
$$\int_{(x,y_0)} f(x) d\nu_{(x_0,y_0)}(x)= \int_\mathbb{R} f(\phi(\phi^{-1}(x_0)+t),y_0)dt= \int_\mathbb{R} f(\phi(s),y_0)ds=\int_\mathbb{R} f(x,y_0) d\phi_*\mathcal{L}(x),$$
by putting $s=\phi^{-1}(x_0)+t$, with $\phi_*\mathcal{L}$ the image measure under $\phi$ of the Lebesgue measure $\mathcal{L}$.
Now the question is : is $\phi_*\mathcal{L}$ absolutely continuous wrt to $\mathcal{L}$, when $\phi$ is a homeomorphism ? The answer is yes is $\phi$ has some regularity ($C^1$ is okay,  Lipschitz would probably do), but if the homeomorphism is only Holder (or worse), no in general. For an horrible example, consider your favorite measure $\mu$ such that 


*

*$\mu(0,+\infty)=+\infty, \mu(-\infty,0)=+\infty$

*$\mu$ is singular to the Lebesgue measure, 

*$\mu$ has no atom

*$\mu$ has full support in $\mathbb{R}$
Then it can be realised as the image measure $\phi_*\mathcal{L}$ for some $\phi$.
The construction goes as follows: for $t>0$, put $\psi(t)=\mu(0,t)$, and for $t<0$, $\psi(t)=-\mu(t,0)$. $\psi$ is continuous (because no atoms), $\psi$ is strictly increasing (because full support), and the limits are $\pm\infty$ at $\pm \infty$. So it is an homeomorphism of $\mathbb{R}$. Let $\phi$ be its inverse. Then one can check that $\phi_*\mathcal{L}=\mu$.
