$N_{G}(E)/C_{G}(E)$ is the Weyl group of $G$? In the algebraic group $G = \operatorname{PGL}_4(\mathbb{C})$, let $E$ denote the subgroup of elements of order dividing 2 in the diagonal maximal torus; it is generated by the images of the three matrices
$$ 
\begin{bmatrix}
1 & 0 & 0 & 0\\
0 & -1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & -1 \\ 
\end{bmatrix},\quad 
\begin{bmatrix}
1 & 0 & 0\\
0 & -1 & 0 \\
0 & 0  & I_2
\end{bmatrix},\quad
\begin{bmatrix}
-1 & 0 \\
0 & I_3 
\end{bmatrix}.
$$
Direct computation shows that $N_{G}(E)/C_{G}(E) \cong S_4\,$, which is the Weyl group of $G$. Analogous result can be obtained for $n = 3, 5, 6, 7, 8, 9, 10$ in $\operatorname{PGL}_{n}(\mathbb{C})$ by direct calculations. Is this observation true for any $n$? Is there any reference relevant?
 A: $\newcommand{\ZZ}{{\mathcal Z}_G}
\newcommand{\NN}{{\mathcal N}_G}
\newcommand{\zz}{{\mathfrak z}_G}
\newcommand{\Lie}{{\rm Lie\,}}
\renewcommand{\tt}{{\mathfrak t}}
\renewcommand{\gg}{{\mathfrak g}}
\newcommand{\X}{{\sf X}}
\newcommand{\Z}{{\Bbb Z}}$
Yes, this is true for any $n\ge 3$.
Let $G$ be a semisimple group of adjoint type
over an algebraically closed field $k$ of characteristic 0.
Let $T\subset G$ be a maximal torus.
Write
$$E=T^{(2)}:= \{t\in T(k)\ |\ t^2=1\}.$$
We wish to compute the centralizer $\ZZ(E)$ and the normalizer $\NN(E)$.
Observe that  $\NN(E)\supseteq \ZZ(E)\supseteq T$.
We compute  $\zz(E):=\Lie \ZZ(E)$.

Lemma 1. $\zz(E)=\tt:=\Lie T$.

Proof.
Since $\ZZ(E)\supseteq T$, we have $\zz(E)\supseteq\tt$.
Write the root decomposition
$$ \Lie G=\tt\oplus\bigoplus_{\beta\in R}\gg_\beta\,,$$
where $R=R(G,T)\subset \X^*(T)$ is the root system.
Since $\zz(E)\supseteq\tt$, we have
$$ \zz(E)=\tt\oplus\bigoplus_{\beta\in M}\gg_\beta$$
for some subset $M\subseteq R$.
Here
$$M=\{\beta\in R\ |\ \beta(t)=1\ \forall t\in E\}.$$
Let $S\subset R$ be a system of simple roots (a basis of $R$).
Since $G$ is of adjoint type, the set $S\subset R\subset \X^*(T)$
is a basis of the character group $\X^*(T)$ of $T$.
It follows that for any simple root $\alpha\in S$,
there exists $t\in E=T^{(2)}$ such that $\alpha(t)=-1$.
Let $W=W(G,T)=\NN(T)/T$ be the Weyl group.
The group $W$ acts on $E$ and on $R$, and $W\cdot S=R$.
Therefore, for any root $\beta\in R$, there exists $t\in E$
such that $\beta(t)=-1$, and therefore
$M=\varnothing$ and $\zz(E)=\tt$, as required.
We compute $\NN(E)$. Since ${\rm char}(k)=0$,
it follows from Lemma 1 that the identity component  $\ZZ(E)^0$ of $\ZZ(E)$ is $T$.
We see that  $\NN(E)$ normalizes $\ZZ(E)$, and hence it normalizes $\ZZ(E)^0=T$.
It follows that  $\NN(E)\subseteq \NN(T)$.
On the other hand, $\NN(T)$ normalizes $T$, and therefore, it normalizes $E=T^{(2)}$,
whence $\NN(T)\subseteq \NN(E)$.
Thus $\NN(E)=\NN(T)$.
We wish to compute $\ZZ(E)$.
Consider the Weyl group  $W=\NN(T)/T$.
Set $$W_E=\ZZ(E)/T\subseteq \NN(T)/T=W.$$
Then $W_E$ is a finite group, the kernel of the natural homomorphism
$W\to{\rm Aut\,} E$,
and $\ZZ(E)$ is the preimage of $W_E$ in $\NN(T)$.

Lemma 2. Let $W'=S_n$ (the permutation group on $n$ symbols)
naturally acting on the set
$$E'=\ker\, \Sigma\colon (\Z/2\Z)^n\to \Z/2\Z,$$
where $\Sigma(x_1,x_2,\dots,x_n)=x_1+x_2+\cdots+x_n\,.$
If $n\ge 3$, then this action is effective (has trivial kernel).

Proof. Left to OP and the reader.
Observe that Lemma 2 is false for $n=2$, when ${\rm Aut\,} E'=\{1\}$
whereas $S_2\neq \{1\}$.

Theorem. Let  $G={\rm PGL}_n$ with $n\ge 3$.
Then $\NN(E)/\ZZ(E)=\NN(T)/T=W$.

Proof.
We have seen that $\NN(E)=\NN(T)$. By Lemma 2,
if $n\ge 3$, then $W_E=\{1\}$, whence $\ZZ(E)=T$, and the theorem follows.
Edit. Observe that the analogues of Lemma 2 and the theorem do not hold for the adjoint group  ${\rm SO}(2n+1)$ of type ${\sf B}_n$ for $n\ge 1$.
Indeed, then
$$W\simeq(\Z/2\Z)^n\rtimes S_n\,,$$
and the normal subgroup $(\Z/2\Z)^n\subseteq W$ acts on $E=T^{(2)}\cong (\pm1)^n$ trivially.
Therefore, $\ZZ(E)\neq T$.
