Isotropy algebra for $U(n)$?

Question

Let $G = U(n)$ be the Liegroup of $n \times n$ unitary matrices and $\mathfrak{g}$ the corresponding Lie algebra. Now $G$ can act on $\mathfrak{g}$ by the Adjoint-action. Since $G$ is a subgroup of $\mathbb{C}^{n \times n}$ the adjoint-action is just

$$G \times \mathfrak{g} \to \mathfrak{g}, \ (X, A) \mapsto XAX^{-1}=XA\overline{X}^T.$$

I know, that $G_A = \{X \in G | XAX^{-1} =A\}$. Is then the Lie algebra of $G_A$ given by $\mathfrak{g}_A = \{X \in \mathfrak{g} | [X,A] = XA-AX =0\}$?

Answer 1

If $U$ is in the Lie algebra of $G_A$, you have $exp(tU)Aexp(-tU)=A$. If you differentiate this, you obtain:

$UA-AU=0$.