Orbits, linear orbits, and invariant subspaces

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Dear mathematical hivemind,

I'd be grateful for any leads on the following questions:

Let $\mathcal{D}$ be a completely reducible (projective) representation of a connected Lie group $G$ on a Hilbert space $\mathcal{H}$. Suppose that $\mathcal{H}$ has the direct sum decomposition into invariant subspaces $\mathcal{H}= \mathcal{H}_1 \oplus \mathcal{H}_2 \oplus...$ where the restrictions of $\mathcal{D}$ to the $\mathcal{H}_i$ are irreducible. The $G$-orbit of a vector $\phi$ in $\mathcal{H}$ is defined as the set $$ \mathscr{O}_G(\phi) = \{\mathcal{D}(g)\phi \; |\; g \in G\}, $$ and the linear $G$-orbit of $\phi$ is the linear closure of $\mathscr{O}_G(\phi)$ with respect to the complex numbers. [Note that, for any $\phi \in \mathcal{H}$, the linear $G$-orbit of $\phi$ is equal to one of the $G$-invariant subspaces of $\mathcal{H}$.]

If $G=SU(2)$, then the following holds: the two-dimensional invariant subspace $\mathcal{H}_{1/2}$ (sometimes called the "spin-1/2 representation") is equal to the $SU(2)$-orbit of any vector in $\mathcal{H}_{1/2}$; i.e. $\mathcal{H}_{1/2} = \mathscr{O}_{SU(2)}(\phi)$ for all $\phi \in \mathcal{H}_{1/2}$. In other words: for any $\phi \in \mathcal{H}_{1/2}$, the $SU(2)$-orbit of $\phi$ is equal to the linear $SU(2)$-orbit of $\phi$.

More generally, consider the following claim: for all $\mathcal{H}_i$, $\mathcal{H}_i = \mathscr{O}_G(\phi)$ for all $\phi \in \mathcal{H}_i$.

Question 1: Does this claim hold in general, i.e. for all connected Lie groups? If so, what's the proof? If not, what's a counterexample?

Question 2: Does this claim hold for the Poincaré group? If so, what's the proof? If not, what's a counterexample?

Question 3: Does this claim hold for $SU(2)$-invariant subspaces of higher dimensions, i.e. dimensions $\geq3$?

Looking forward to your responses!