Let $G$ be a compact connected Lie group with Lie algebra ${\frak g}$. For $\lambda\in{\frak g}^*$ let $$G_\lambda=\{g\in G:{\rm Ad}_g^*\lambda=\lambda\},$$ i.e. $G_\lambda$ is the stabilizer of $\lambda$ for the coadjoint action of $G$ on ${\frak g}^*$.
Question: When does $G_\lambda=G_\mu$?
But maybe this is a too vague question. Let me be more precise (I look for a proof of the Proposition bellow):
Let $T$ be a maximal torus with Lie algebra ${\frak t}$. Then, ${\frak t}$ determines a root system and we let ${\cal C}\subseteq{\frak t}^*$ be the closed fundamental Weyl chamber. The root decomposition is ${\frak g}={\frak t}\oplus{\frak b}$ for some ${\frak b}$, so we can view ${\frak t}^*$ as a subset of ${\frak g}^*$ by identifying it with the annihilator of ${\frak b}$. Also, recall that the closed Weyl chamber ${\cal C}$ is a disjoint union of $2^r$ open faces (sometimes called wall), where $r$ is the rank of ${\frak g}$.
Proposition: If $\lambda,\mu\in{\cal C}$ Lie in the same face of ${\cal C}$, then $G_\lambda=G_\mu$.
It is mentioned, for example, here, but they don't say why. Have you any idea?
Here is my idea: As $G_{\lambda}, G_{\mu}$ are both closed subgroups of $G$, they are uniquely determined by their lie algebras. The Lie algebra of $G_{\lambda}$ is given by $$\mathfrak{g}_{\lambda} = \{a \in \mathfrak{g}: \lambda([a, \mathfrak{g}]) = 0\}.$$ We now compute this. Assume $a$ lies in $\mathfrak{g}_{\lambda}$, with $a = t + b$ for $t \in \mathfrak{t}$ and $b \in \mathfrak{b}$. Then as $[\mathfrak{t}, \mathfrak{g}] \subset \mathfrak{b}$, we have $\lambda([a, \mathfrak{g}]) = \lambda([b, \mathfrak{b}])$. We note that the projection of $[b, \mathfrak{b}]$ onto $\mathfrak{t}$ is the span of all roots $h$ such that the projection of $b$ in $V_h$ is non-zero(We identify $\mathfrak{t}$ with $\mathfrak{t}^*$ here). Therefore, we conclude that $$\mathfrak{g}_{\lambda} = \mathfrak{t} \bigoplus_{t \in \mathfrak{t}^*: \langle t, \lambda \rangle = 0} V_t.$$ The sum depends only on which face $\lambda$ is on, so we have $G_{\lambda} = G_{\mu}$ as desired.