Is equivariant immersion by parts w.r.t an action with finitely many orbits an immersion? Let $M,N$ be a smooth manifolds of dimension greater than $2$. Suppose that there is a Lie group $G$ acting on $M,N$, and that $f:M \to N$ is a smooth injective equivariant map. 
Suppose further that the action of $G$ on $M$ has a finite number of orbits $H_i$ , which are embedded submanifolds of $M$. Since $f$ is equivariant, its restriction $f|_{H_i}$ has a constant degree on each orbit $H_i$. Since 
it is injective, all the $f|_{H_i}$ are immersions (by the constant rank theorem). 
Question: Is it true that $f:M \to N$ is an immersion?
In general, an "immersion by parts" does not need to be an immersion; The dimensions of the orbits $H_i$ can differ, and the "total rank" can fall when passing between the different $H_i$.
However, I wonder if there is a chance for something like this to hold, when we assume there is a finite number of orbits for the action.
 A: Not really. Fix some $n \in \mathbb{N}$ and let $G = \prod_{i=1}^n GL_1(\mathbb{R})$. Set $M = N = \mathbb{R}^n$ and consider the following two actions of $G$:


*

*The group $G$ will act on $M$ by $(a_1, \dots, a_n) \cdot (x_1, \dots, x_n) = (a_1 x_1, \dots, a_n x_n)$.

*The group $G$ will act on $N$ by $(a_1, \dots, a_n) * (x_1, \dots, x_n) = (a_1^3 x_1, \dots, a_n^3 x_n)$. 


Those are readily seen to be group actions and the action of $G$ on $M$ has $2^n$ orbits since given a subset $A \subseteq \{ 1, \dots, n \}$, all the vectors $(x_1,\dots,x_n)$ such that $x_i = 0$ if $i \in A$ and $x_i \neq 0$ if $i \notin A$ belong to the same orbit. Those orbits are clearly embedded submanifolds. The map $f \colon M \rightarrow N$ given by $f(x_1, \dots, x_n) = (x_1^3, \dots, x_n^3)$ is smooth, injective and equivariant with respect to the actions above since
$$ f((a_1, \dots, a_n) \cdot (x_1, \dots, x_n)) = f(a_1x_1, \dots, a_nx_n) = ((a_1x_1)^3, \dots, (a_nx_n)^3) = (a_1^3 x_1^3, \dots, a_n^3 x_n^3) = (a_1, \dots, a_n) * (x_1, \dots, x_n). $$
However, on the orbit described by the subset $A$, it has rank $n - |A|$.
