Suppose an connected algebraic group $G$ defined over $\mathbb{F}_q$ acts on an algebraic variety $V$ defined over $\mathbb{F}_q$ by an action defined over $\mathbb{F}_q$. If $F$ is a Frobenius endomorphism on $G$, then any $F$-stable $G$-orbit contains a rational point.
Proof: If $v$ is in an $F$-stable orbit, then $F\cdot v=g\cdot v$ for some $g\in g$. By Lang Steinberg, $g=F(h)^{-1}h$ for some $h\in G$. Then $F\cdot v=(F(h)^{-1}h)\cdot v$. This implies $F(h)\cdot(F\cdot v)=h\cdot v$, i.e., $F\cdot(h\cdot v)=h\cdot v$, so $h\cdot v$ is a rational point in the orbit of $v$.
Why does $F(h)\cdot(F\cdot v)=F\cdot(h\cdot v)$? Is this some assumption on the action of $F$? I'm confused because $F$ is technically a function on $G$, but $F$ and $G$ both act on $V$ somehow. Is there some sort of assumed compatibility between these actions?