Everything I will say is over complex numbers. Let $G$ be an algebraic subgroup of $GL(V)$ (not necessarily connected) and let $G_0$ be the connected component of $G$ that contains the identity. $G_0$ is normal in $G$ and there is an exact sequence $$ 1 \to G_0 \to G \to \Gamma \to 1 $$ where $\Gamma = G/G_0$ is a finite group (because it is algebraic and discrete - is this correct?).
I am looking for sufficient conditions that guarantee that this sequence splits, providing $G = G_0 \rtimes \Gamma$. We can restrict to the case where $G_0$ is reductive (or even semisimple) if needed.
I know that in general the sequence does not split (even if $G_0$ is reductive): see for instance Does the quotient of an algebraic group by its neutral component always split?. This example generalizes as follows: let $G_0$ be the "$SL$-torus" of diagonal matrices with determinant $1$ and $G = N_{SL(V)}(G_0)$ so that $\Gamma = \mathfrak{S_n}$ (if $\dim V = n$) is the Weyl group of $SL(V)$. The sequence does not split in this case and the reason is that a splitting map would embed $\mathfrak{S_n}$ into $SL(V)$, but this is not possible. I hope I understood the example correctly.
I am wondering whether the sequence splits if one considers the case where $G = N_{GL(V)}(G_0)$. At least in this case, the natural generalization of the example above is no longer a counterexample, since the normalizer of the $GL$-torus $T$ of invertible diagonal matrices is indeed $T \rtimes \mathfrak{S_n}$.