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Consider the Lie group $SU(2)$. A maximal torus for $SU(2)$ is $$T=\left\{\begin{pmatrix}e^{i\theta} & 0 \\ 0 & e^{-i\theta}\end{pmatrix}:\theta\in{\Bbb R}\right\},$$ and its Lie algebra is the maximal abelian subalgebra of ${\frak su}(2)$ $${\frak t}={\rm span}(\sigma),\quad\text{where}\quad\sigma:=\begin{pmatrix}i & 0 \\ 0 & -i \end{pmatrix}.$$ In this particular case, it is easy to verify directly that the maximal torus $T$ is the set of matrices that fixes its Lie algebra by the adjoint action, i.e. $$ \begin{align} T &= \{g\in SU(2):g\sigma g^{-1}=\sigma\} \\ &= \{g\in SU(2):{\rm Ad}_g(X)=X\text{ for all }X\in{\frak t}\}. \end{align} $$

Question: Does this generalizes to every compact connect Lie group $G$?

In other words, let $G$ be a compact connected Lie group with maximal torus $T$ and corresponding maximal abelian subalgebra ${\frak t}$. Is it true that $$T=\{g\in G:{\rm Ad}_g(X)=X\text{ for all }X\in{\frak t}\}?\tag{1}$$

Or is there a condition on the Lie group such that this condition holds? For example, is it true for $SU(n)$?

Edit: I am able to show one inclusion. Let $G_{\frak t}$ be the right-hand side of $(1)$. Then, we have $T\subseteq G_{\frak t}$. Indeed, let $X\in{\frak t}$. Then, for all $Y\in{\frak t}$, $$\frac{d}{dt}\Big|_{t=t_0}{\rm Ad}_{\exp tX}Y={\rm Ad}_{\exp t_0X}[X,Y]=0,$$ since ${\frak t}$ is abelian. Thus, ${\rm Ad}_{\exp X}Y=Y$ for all $X,Y\in{\frak t}$. Since $T$ is connected, it follows that $T\subseteq G_{\frak t}$.

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For all $g, X$, $\exp Ad_g X= g \exp X g^{-1}$ : in fact $f(t)=\exp Ad_g tX$ is a one parameter subgroup of $G$ with $f'(0)=Ad_gX$. Thus if $Ad_g $ induces the identity on $t$, the element $g$ commute with $T$, but $T$ is a maximal abelian subgroup, and $g\in T$.

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