Is the image of an equivariant map always a weakly embedded submanifold? Let $M,N$ be smooth manifolds, with a smooth $G$-action on them, by some Lie group $G$. Suppose also that $M$ has a finite number of orbits under $G$'s-action. 
Let $f:M \to N$ be a smooth, equivariant, injective immersion.

Is $f(M)$ a weakly embedded submanifold of $N$?

Weakly embedded here means that for every manifold $Q$ and for every smooth map $h:Q \to N$, with $h(Q)\subset f(M)$,the associated map $h:Q\to f(M)$ is also smooth. In other words, it's always valid to restrict the range. 
It is known that it suffices to prove that $h:Q\to f(M)$ is continuous. Note that in general $f(M)$ is only an immersed submanifold. In particular, it can have more open sets than those that come from the subspace topology.
Weakly embedded is a notion which is between "immersed" and "embedded". It is also known that every Lie subgroup is weakly embedded.

A famous example for a weakly embedded submanifold, which is not embedded is the dense curve on the torus. (In that case, there is also a Lie group action in the background, by $\mathbb{R}$).
 A: Here is a counter-example with $G = (\mathbb{R}, +)$, $M = \mathbb{R}$ and $N = \mathbb{R}^2$. I shall specify the actions later on.
Consider on $N$ (equipped with cartesian coordinates $(u,v)$) the smooth function $h(u,v) = v(u^4 + v^4 + u^2 - v^2)$. Here is a figure of some of its level sets. We notice in particular that there is a figure 8 inside the level $[h=0]$; we shall denote this subset $8$. To be safe, add to $h$ a smooth nonnegative exhausting function which vanishes identically on a ball which contains $8$ (so as to leave unchanged the local picture), but which grows sufficiently fast to ensure that the resulting function (call it also $h$) only has compact level sets.
Consider now the (smooth) vector field $X$ on $N$ which is the symplectic gradient of $h$, that is the (usual) gradient vector field everywhere rotated (clockwise, say) by 90°. The field $X$ is everywhere tangent to the levels sets of $h$. The flow of $X$ turns out to be complete (since the levels sets are compact and $X$ smooth) and smooth; it determines a (smooth) $G$-action $\alpha_N : G \times N \to N : (g, (u,v)) \mapsto \Phi^{g}_{X}(u,v)$.
Let $f : M \to N$ be any smooth injective immersion whose image is $8$; it is not a weak embedding. We can pullback $X$ to $M$ along $f$ to get a (smooth and complete) vector field $Y$ which turns out to vanish in only finitely many points. Consider the $G$-action $\alpha_M : G \times M \to M : (g, x) \mapsto \Phi^g_Y(x)$.
By construction, $\alpha_M$ has only finitely many orbits and $f$ is $G$-equivariant since $X$ and $Y$ are $f$-related.
