@LSpice has already proven my revised conjecture in the updated answer to https://mathoverflow.net/questions/377981/classification-of-not-necessarily-connected-compact-lie-groups?noredirect=1#comment959636_377981, but let me give another, closely related proof.

Since $1\to \mathrm{Inn}(G_0) \to \mathrm{Aut}(G_0) \to \mathrm{Out}(G_0) \to 1$ always splits, see https://mathoverflow.net/questions/378218/, we can choose a subgroup $R_0 \subseteq \mathrm{Aut}(G_0)$ for which the restriction of $\mathrm{Aut}(G_0) \to \mathrm{Out}(G_0)$ is an isomorphism. The inverse image of $R_0$ under the map $f:G \to \mathrm{Aut}(G_0)$ induced by conjugation is a subgroup $K \subseteq G$ whose intersection with $G_0$ is $Z(G_0)$.

 Multiplying any $g\in G$ by arbitrary $h \in G_0$ multiplies the associated $f(g) \in \mathrm{Aut}(G_0)$ by an arbitrary inner automorphism $f(h) \in \mathrm{Inn}(G_0)$, without changing $g$'s connected component. Thus, $K$ meets every connected component of $G$.

Using the result of https://mathoverflow.net/questions/150949, $K$ has a finite subgroup $R$ that meets every component of $K$, hence it meets every component of $G$ as well, and intersects $G_0$ within $Z(G_0)$. By design, the elements of $R$ either act by non-trivial outer automorphisms on $G_0$ or they act trivially on $G_0$. This proves my (revised) conjecture.