This question may be trivial to people with the right background, but I do not see the answer.

Let $\Bbbk$ be an algebraically closed field. Can any one-dimensional group variety (over $\Bbbk$) act transitively on $\mathbb{P}^1_{\Bbbk}$?

Here is my reasoning so far:

-Every $g \in G$ fixes a point of $\mathbb{P}^1$ since the associated linear map has an eigenvector.

-Let $G_0$ be the identity component of $G$. If $G_0$ fixes a point $p$, then $G/G_0$ surjects onto the orbit of $p$, so $p$ has finite $G$-orbit. Hence $G$ does not act transitively. So, it suffices to assume $G$ is connected and show that it fixes a point.

-I think (but am not confident here) that the only one-dimensional connected group varieties are $(\Bbbk, +)$, $(\Bbbk^*, \cdot)$, and abelian varieties (i.e., elliptic curves). The last are not an issue since any morphism $E \to PGL_2$ is constant, where $E$ is complete and $PGL_2$ is affine.

-Since $G$ is one-dimensional irreducible and the isotropy subgroup of a point is closed, to show that $G$ fixes $p$, it suffices to show that infinitely many elements of $G$ fix $p$.

The last step, I think I can do for the multiplicative group, and for the additive group if $\Bbbk$ has characteristic zero, but my basic line of approach does not seem to work for the additive group in characteristic $p$. I also am less than completely confident of the classification I've stated for one-dimensional connected group varieties; if someone could point me to a good reference for this, I would appreciate it.