Let $G$ be a compact connected semisimple Lie group, $\mathfrak{g}$ be its complexified Lie algebra and $\mathfrak{g}^*$ its complex dual space. We can form the symmetric algebra $S(\mathfrak{g}^ * ) $ and its $G$-invariant subalgebra $S(\mathfrak{g}^ * ) ^G$. For a closed connected subgroup $H< G$ we have the same construction and get $S(\mathfrak{h}^ * )^H$.

Since $\mathfrak{h} \hookrightarrow \mathfrak{g}$ we get $\mathfrak{g}^*\twoheadrightarrow \mathfrak{h}^ * $ hence $S(\mathfrak{g}^ * ) \twoheadrightarrow S(\mathfrak{h}^ * )$. Take invariant subalgebra we get the map $$ \phi: S(\mathfrak{g}^ * )^G \rightarrow S(\mathfrak{h}^ * )^H. $$

Notice that $\phi$ is not always a projection map. For example when $H=T$ is a Cartan subalgebra, then it is well-known that $S(\mathfrak{g}^ * )^G \cong S(\mathfrak{t}^ * )^W$ is the invariant subalgebra under the Weyl group action and $S(\mathfrak{t}^ * )^T=S(\mathfrak{t}^ * )$ since $T$ is abelian. Therefore $\phi: S(\mathfrak{g}^ * )^G \rightarrow S(\mathfrak{t}^ * )^T$ is the embedding (injective but not surjective).

My question is: for which $H$, the map $\phi$ is injective? We know that $H=G$ itself or $H$ is a Cartan subgroup makes $\phi$ an injection. Are these the only cases?