This is probably a very simple question with a negative answer, but I somehow cannot find a counterexample.
Let $X$ be a smooth algebraic variety over an algebraically closed field $k$. Assume that $X$ contains an embedded curve $C$ isomorphic to $\mathbb{A}^1_*$, such that for every $\varphi\in \operatorname{Aut}(X)$ we have $\varphi(C)=C$; and the induced action of $\operatorname{Aut}(X)$ on $C$ is transitive. Is it true that $\operatorname{Aut}(X)$ contains a subgroup isomorphic to $\mathbb{G}_{m}$, acting transitively on $C$?
Equivalently, let $\operatorname{Aut}(X)^{\circ}$ be the identity component of $\operatorname{Aut}(X)$, and let $\operatorname{Stab}(C)^{\circ}$ be the subgroup of the connected component of $\operatorname{Aut}(X)^{\circ}$ consisting of those automorphisms which fix $C$ pointwise. I assume that the restriction map $\operatorname{Aut}(X)^{\circ}\to \operatorname{Aut}(C)^{\circ}\cong \mathbb{G}_{m}$ is surjective, and I ask whether the exact sequence $1\to \operatorname{Stab}(C)^{\circ} \to \operatorname{Aut}(X)^{\circ} \to \mathbb{G}_{m}\to 1$ splits on the right. Put this way, it seems false since extensions of groups are usually not semidirect products. But maybe with $\mathbb{G}_{m}$ this is true?
On the other hand, I have the following "geometric" intuition. Assume $k=\mathbb{C}$. Since $C\hookrightarrow X$ is an embedding, it should give an injective map between tangent bundles, and so, between global vector fields. This means that I should have a vector field on $X$ tangent to $C$, whose flow on $C$ is my $\mathbb{C}^{*}$-action. Integrating this vector field, I should get a holomorphic $\mathbb{C}^{1}$- or $\mathbb{C}^{*}$-action on $X$, but it cannot be a $\mathbb{C}^{1}$-action because then it would be trivial on $C$. Of course, this action is a priori only holomorphic, but the fact that it restricts to an algebraic action on $C$ suggests that something better might be true.
I would also like to ask an analogous question if $C$ is isomorphic to $\mathbb{A}^{1}$, so I look for a $\mathbb{G}_{a}\subseteq \operatorname{Aut}(X)$ and the same intuition applies; and also if $C$ is a finite set (so I look for some finite subgroup of $\operatorname{Aut}(X)$).
I am not going to think too hard on the geometric problem with the curve, but as for the question about algebraic groups, the answer is negative.
Example: $1\to \mu_n\to \mathbb{G}_m \xrightarrow{x\mapsto x^n} \mathbb{G}_m\to 1$ is not split.