[proofplan] We prove the two implications separately. The Weyl-orbit implication follows directly from the definition of the Weyl [group action](/page/Group%20Action) by choosing a representative in the normalizer. For the converse, we assume $t=gsg^{-1}$ and work inside the identity component of the centralizer of $t$; both $T$ and $gTg^{-1}$ become maximal tori there. The [conjugacy theorem for maximal tori](/theorems/9720) inside that connected compact centralizer then produces an element correcting $g$ into the normalizer of $T$, and this normalizer element sends $s$ to $t$. [/proofplan]

[step:Check that the Weyl action is represented by conjugation in $G$] Let $n\in N_G(T)$. Since $nTn^{-1}=T$, conjugation by $n$ restricts to a map \begin{align*} c_n:T\to T \end{align*} defined by \begin{align*} c_n(x)=nxn^{-1}. \end{align*} If $a\in T$, then $na$ represents the same coset as $n$ in $N_G(T)/T$, and for every $x\in T$, \begin{align*} (na)x(na)^{-1}=n(axa^{-1})n^{-1}=nxn^{-1}, \end{align*} because $T$ is abelian. Hence the formula \begin{align*} (nT)\cdot x=nxn^{-1} \end{align*} is well-defined. Suppose first that $s$ and $t$ lie in the same $W(G,T)$-orbit. Then there is a coset $nT\in W(G,T)$ with \begin{align*} t=(nT)\cdot s=nsn^{-1}. \end{align*} Since $n\in N_G(T)\le G$, this equation says exactly that $s$ and $t$ are conjugate in $G$. [/step]

[step:Place the two tori inside the identity component of the centralizer of $t$]Assume conversely that $s$ and $t$ are conjugate in $G$. Choose $g\in G$ such that \begin{align*} t=gsg^{-1}. \end{align*} Define the centralizer of $t$ in $G$ by \begin{align*} C_G(t):=\{x\in G:xt=tx\}, \end{align*} and let \begin{align*} H:=C_G(t)^\circ \end{align*} denote its identity component. Since $t\in T$ and $T$ is abelian, every $a\in T$ satisfies $at=ta$, so $T\le C_G(t)$. The torus $T$ is connected and contains the identity element $e$, hence $T\le H$. Similarly, let $b\in gTg^{-1}$. Then there is $a\in T$ such that $b=gag^{-1}$. Since $T$ is abelian, $as=sa$, and therefore \begin{align*} bt=(gag^{-1})(gsg^{-1})=gasg^{-1}=gsag^{-1}=(gsg^{-1})(gag^{-1})=tb. \end{align*} Thus $gTg^{-1}\le C_G(t)$. Since $gTg^{-1}$ is connected and contains $e$, it follows that \begin{align*} gTg^{-1}\le H. \end{align*}[/step]

[guided]We now move the problem into a smaller group where the usual conjugacy theorem for maximal tori can be applied. The element $t$ is fixed throughout this part, so define \begin{align*} C_G(t):=\{x\in G:xt=tx\} \end{align*} and let \begin{align*} H:=C_G(t)^\circ \end{align*} be the [connected component](/page/Connected%20Component) of $C_G(t)$ containing the identity element $e$. First we verify that $T$ lies in $H$. Because $t\in T$ and $T$ is abelian, every $a\in T$ commutes with $t$: \begin{align*} at=ta. \end{align*} Hence $T\le C_G(t)$. The subgroup $T$ is connected and contains $e$, so it must lie in the identity component of $C_G(t)$; therefore $T\le H$. Next we verify the same statement for the conjugate torus $gTg^{-1}$. Take an arbitrary element $b\in gTg^{-1}$. By definition of the conjugate subgroup, there exists $a\in T$ such that \begin{align*} b=gag^{-1}. \end{align*} Using the assumption $t=gsg^{-1}$ and the commutativity relation $as=sa$ inside $T$, we compute \begin{align*} bt=(gag^{-1})(gsg^{-1})=gasg^{-1}=gsag^{-1}=(gsg^{-1})(gag^{-1})=tb. \end{align*} Thus every element of $gTg^{-1}$ commutes with $t$, so $gTg^{-1}\le C_G(t)$. The subgroup $gTg^{-1}$ is connected because it is the image of the connected group $T$ under the homeomorphism $x\mapsto gxg^{-1}$, and it contains $e$. Hence it lies in the identity component: \begin{align*} gTg^{-1}\le H. \end{align*}[/guided]

[step:Show that both tori are maximal inside the connected centralizer] The subgroup $C_G(t)$ is closed in $G$, because it is the preimage of the diagonal under the continuous map \begin{align*} G\to G\times G \end{align*} given by \begin{align*} x\mapsto (xt,tx). \end{align*} Therefore $C_G(t)$ is compact, and its identity component $H=C_G(t)^\circ$ is a compact connected Lie group. We show that $T$ is a maximal torus of $H$. Let $S\le H$ be a compact connected abelian Lie subgroup with $T\le S$. Since $H\le G$, the subgroup $S$ is also a compact connected abelian Lie subgroup of $G$ containing $T$. By maximality of $T$ as a torus in $G$, we get $S=T$. The conjugate subgroup $gTg^{-1}$ is also a maximal torus of $G$, because conjugation by $g$ is a Lie group automorphism of $G$ carrying compact connected abelian Lie subgroups to compact connected abelian Lie subgroups and preserving inclusions. If $S'\le H$ is a compact connected abelian Lie subgroup with $gTg^{-1}\le S'$, then $S'$ is such a subgroup of $G$ containing the maximal torus $gTg^{-1}$, so \begin{align*} S'=gTg^{-1}. \end{align*} Thus $T$ and $gTg^{-1}$ are maximal tori of $H$. [/step]

[step:Conjugate the two maximal tori inside the centralizer] Since $H$ is a compact connected Lie group and $T$ and $gTg^{-1}$ are maximal tori of $H$, the [Conjugacy Theorem for Maximal Tori][citetheorem:9720] gives an element $h\in H$ such that \begin{align*} h(gTg^{-1})h^{-1}=T. \end{align*} Because $H\le C_G(t)$, the element $h$ commutes with $t$. [/step]

[step:Produce a normalizer representative sending $s$ to $t$] Define \begin{align*} n:=hg\in G. \end{align*} From \begin{align*} h(gTg^{-1})h^{-1}=T \end{align*} we obtain \begin{align*} nTn^{-1}=T, \end{align*} so $n\in N_G(T)$. Moreover, using $t=gsg^{-1}$ and $h\in C_G(t)$, \begin{align*} nsn^{-1}=h(gsg^{-1})h^{-1}=hth^{-1}=t. \end{align*} Therefore $t=(nT)\cdot s$, so $s$ and $t$ lie in the same $W(G,T)$-orbit. Combining this with the first implication proves the equivalence. [/step]