This question is a follow-up to Classification of (not necessarily connected) compact Lie groups. In the answer to that question, @LSpice proved that any compact, not necessarily connected Lie group $G$ takes the form $$ G = \frac{G_0 \rtimes R}{P} $$ where $G_0$ is the identity component of $G$, $R$ is a finite group, and $P$ is finite and central.

Nonetheless, there are many possibilities for the semi-direct product. To narrow the list, it would be convenient to separate out those elements of $R$ that act by non-trivial outer automorphisms on $G_0$ and modify the rest so that they commute with $G_0$. The simplest possibly-correct realization of this is:

UPDATE: my original hypothesis (below) is false. A weaker, possibly correct version is:

Hypothesis: $R$ and $P$ can be chosen above such that every element of $R$ either (1) acts by a non-trivial outer automorphism on $G_0$ or (2) acts trivially on $G_0$.

I suspect that this can be proven using the answer to Does Aut(G) → Out(G) always split for a compact, connected Lie group G?, together with In any Lie group with finitely many connected components, does there exist a finite subgroup which meets every component?.

By comparison, this is false:

Hypothesis: Any compact Lie group $G$ can be written in the form $$ G = \frac{(G_0 \times H) \rtimes R}{P} $$ where $H, R, P$ are finite groups, non-trivial elements of $R$ act by non-trivial outer automorphisms on $G_0$, and $P$ is central.

Counterexample: consider $G = U(1) \rtimes \mathbb{Z}_4$, where the generator $r$ of $\mathbb{Z}_4$ acts by the ``charge conjugation'' outer automorphism $r^{-1} e^{i \theta} r = e^{-i \theta}$ on $U(1)$. In any finite central extension $G'$ of this group, elements of $\pi_0(G)$ that act by charge conjugation will never square to the identity in $G'$, so $G'$ never takes the required $(G\times H) \rtimes \mathbb{Z}_2$ form with $\mathbb{Z}_2$ acting on $U(1)$ by charge conjugation.