On the orbits of an action of the subgroup of rational points Let $k$ be a field, $G$ an affine algebraic group over $k$ and $V$ an irreducible affine variety over $k$. Suppose that $G$ acts $k-$morphically on $V$ and that the action is given by $(g,x) \mapsto g \cdot x$ where $g \in G, x \in V$. In particular, the group $G(k)$ of rational $k-$points acts on the set $V(k)$ of $k-$rational points of $V$. 
I am wondering if the following assertion holds:

$$G(k) \cdot x = (G \cdot x) \cap V(k) \;\; \text{for each} \; x \in V(k).$$ 

That is to say, the $G(k)-$orbit of a $k-$rational point of $V$ consists of the $k-$rational points in its $G-$orbit.
Remarks. 
(i) The inclusion $G(k) \cdot x \subseteq (G \cdot x) \cap V(k)$ should be clear.
(ii) If $V(k)= \emptyset$ then the statement is vacuously true, so we may assume that $V(k) \neq \emptyset$.
(iii) The above statement holds in the following example: $G=\textbf{G}_a=\overline{k}^{\;+}$ acting by translations on $V=\mathbb{A}_{\; \overline{k}}^n$ (in this case $G(k)=k^+$ acts by translations on $V(k)=\mathbb{A}_k^n$). 
Actually, I am satisfied with a positive answer in the case where $V=\mathbb{A}^n_{\;\overline{k}}$ is the $n-$dimensional affine space over $\overline{k}$ (the algebraic closure of $k$). Thank you in advance.
 A: This is false even when $V=\mathbb{A}^1_\mathbb{Q}$. For exmample, let $\mathbb{G}_{m,\mathbb{Q}}$ act on $V$ as follows: for a $\mathbb{Q}$-algebra $R$ and an element $r\in\mathbb{A}^1_\mathbb{Q}(R)$ and $g\in\mathbb{G}_{m,\mathbb{Q}}(R)=R^\times$ we set $g\cdot r=g^2 r$. 
Let us then note the following. If we take $1\in\mathbb{A}^1_\mathbb{Q}(\mathbb{Q})$ then $\mathbb{G}_{m,\mathbb{Q}}(\mathbb{Q})\cdot 1=\{g^2:g\in\mathbb{Q}^\times\}\subsetneq \mathbb{Q}^\times$. We claim that the scheme-theoretic orbit $O_1$ of $\mathbb{G}_{m,\mathbb{Q}}$ acting on $1$ is all of $\mathbb{G}_m$. The reason is that since $O_1\subseteq \mathbb{G}_{m,\mathbb{Q}}\subseteq\mathbb{A}^1_\mathbb{Q}$ is locally closed it suffices to show that $O_1(\overline{\mathbb{Q}})=\mathbb{G}_{m,\mathbb{Q}}(\overline{\mathbb{Q}})=\overline{\mathbb{Q}}^\times$. But this is clear since every element in $\overline{\mathbb{Q}}^\times$ is a square. So, $O_1(\mathbb{Q})=\mathbb{Q}^\times$.
So we have a proper containment $\mathbb{G}_{m,\mathbb{Q}}(\mathbb{Q})\cdot 1\subsetneq O_1(\mathbb{Q})$. 
As pointed out by Brian in the linked post one can understand the discrepancy in terms of a cohomology group. Unfortunately what this cohomology group is can be quite complicated. If we're in characteristic $0$ it's much simpler. Namely, if $G$ acts on $V$ and $x\in V(k)$ then one can understand the discrepancy between $G(k)\cdot x$ and $O_x(k)$ as being measured by $H^1_\text{cont.}(\mathrm{Gal}(\overline{k}/k),G_x(\overline{k}))$ where $G_x$ is the isotropy subgroup of $G$ corrresponding to $x$. In our case this isotropy subgroup is $\mu_2$ and as you might know from Kummer theory one has that $H^1_\text{cont.}(\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}),\mu_2)=\mathbb{Q}^\times/(\mathbb{Q}^\times)^2$ which is precisely the obstruction we observed by working things out by hand.
The reason it had to work with $\mathbb{G}_a$ is essentially the following: in characteristic $0$ the only subgroups of $\mathbb{G}_a$ are itself and the trivial subgroup and both of those have trivial cohomology groups. It actually might be an issue in characteristic $p$ if you take the action $g\cdot x:=x+g^p$.
