Let $G$ be a connected algebraic group, $X$ a variety and and $\theta : G\to X$ be a morphism of varieties. (In particular it could be the orbit of some action of $G$.)
Consider the corresponding differential $d\theta : T_eG \to T_{\theta(e)}X$. (Where $T_eG$ is the corresponding Lie algebra.)
Is it true that $\theta$ has dense (hence open) image if and only if $d\theta$ is surjective?
An example would be $G=GL(V)$, $X=V\oplus V$, $v_1,v_2\in V$ where $v_1$ and $v_2$ are not parallel and the map is given by $g\mapsto (g(v_1), g(v_2))$. The image will be dense and consist of all vectors $(v_1,v_2)$ that are not parallel. The corresponding differential sends $M\in \mathrm{End}(V)$ to $(Mv_1,v_2)+(v_1,Mv_2)$ and is surjective. (Choose a matrix $A$ such that $Av_1=x$ and $Av_2=0$ and conversely for $B$ and then $(A+B)(v_1,v_2)=(x,0)+(0,y)=(x,y)$.)
If $d\theta$ is surjective, then $\dim G = \dim T_e G \geq \dim T_{\theta(e)}X \geq \dim X$. So at least if $X$ is connected, then $\theta$ has dense image in $X$. If $X$ is not connected, then clearly this is not the case.
Now say that we are working over a perfect field of characteristic $p > 0$. Consider a map like $\theta: x \mapsto x^p$ from $G = \mathbb{G}_m$ to itself. Then this map is bijective but the differential $d\theta$ is zero. For the other direction to be true, you need to assume something like $\theta$ separable.