An abstract group action need not be an algebraic action

Conjugation is additive, so $g\cdot (h\cdot x)=x+\bar{h}+\bar{g}=(g+h)\cdot x$ and $0\cdot x=x$. Each fixed $g$ acts by the polynomial translation $x\mapsto x+\bar{g}$. But the joint action is not a morphism: restriction along $g\mapsto (g,0)$ would make $g\mapsto \bar{g}$ a polynomial on ${\mathbb{A}}^1$. Such a polynomial would equal $g$ on all real numbers, hence be the identity polynomial, but would then send $i$ to $i$ rather than $-i$.

Inverse pullback acts on $W=\mathbb{C}1+\mathbb{C}z$ by $1\mapsto 1$ and $z\mapsto z-\bar{g}$. Thus $W$ is stable, yet its representation has nonregular matrix coefficient $-\bar{g}$, by step 1.1. No larger finite-dimensional stable subspace containing $z$ can be algebraic: it contains every translate $z-\bar{g}$, in particular the translate $z-1$ for the group element $1$, so it also contains the difference $z-(z-1)=1$ and hence the stable $W$, and restriction to $W$ would have regular matrix entries by taking a finite basis extending a basis of $W$. Consequently the coordinate representation is not rational, even though each polynomial's translates lie in its finite-dimensional bounded-degree space. This explicitly violates the rationality conclusion in F2, and refutes the claim.