[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*}
[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]
[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]
