While trying to understand more the structure of reductive group, I came upon the situation that I describe below. I can't find a mistake in my dissertation, however I arrive to an absurd conclusion. Would somebody agree to read the following and tell me where I assumed something impossible / concluded something wrong ?
Let $G$ be a connected reductive group over an algebraically closed field. Fix $T_0$ a maximal torus, and denote by $W_0:=W(T_0)=N_G(T_0)/T_0$ the associated Weyl group. Assume that there exists an element $w_0\in W_0$ of order $2$ which is not in the center of $W_0$.
Let $n_0\in N_G(T_0)$ be a representative of $w_0$, and consider $\phi:G\xrightarrow{\sim} G$ the conjugation-by-$n_0$ automorphism. Naturally, the torus $T_0$ is $\phi$-stable, and $\phi$ induces the conjugation-by-$w_0$ automorphism of $W_0$.
Let me now consider another maximal torus $T=\,^gT_0$ for some $g\in G$. Associated to it, we have another description of the Weyl group $W:=W(T)=N_G(T)/T \simeq \, ^gW_0$. Note that we also have $N_G(T) = \, ^gN_G(T_0)$.
We make the assumption that $T$ is also $\phi$-stable, so that $\phi$ also induces an action on $W$ naturally. It means that $g^{-1}\phi(g)$ lies in $N_G(T_0)$, I denote this element by $\tilde{n_0}$. Eventually, I assume that $\tilde{n_0}$ also has image $w_0$ in $W_0$. It implies that $\tilde{n_0}n_0$ lies in $T_0$, because $w_0$ has order $2$.
Let me now look at the action of $\phi$ on $W$. Let me consider $\nu\in N_G(T)$ and write $\nu=g\nu_0g^{-1}$ for some $\nu_0\in N_G(T_0)$. I have
$$\begin{align} \phi(\nu)=\phi(g)\phi(\nu_0)\phi(g)^{-1} & = g\tilde{n_0}\,\phi(\nu_0)\,\tilde{n_0}^{-1}g^{-1}\\ & = g\tilde{n_0}\,n_0\nu_0n_0^{-1}\,\tilde{n_0}^{-1}g^{-1}\\ & = (g\tilde{n_0}n_0g^{-1})(g\nu_0g^{-1})(gn_0^{-1}\tilde{n_0}^{-1}g^{-1}) \end{align}$$
Because $\tilde{n_0}n_0$ lies in $T_0$, both the first and last parentheses are in $^gT_0=T$. Moreover, the middle parenthesis is just $\nu$. Now, $\nu$ normalizes $T$ so we may rewrite it under the form $\nu\cdot \text{sth in }T$. This last observation means that $\phi$ acts like $\mathrm{id}$ on $W$. But that is absurd : if it did, it would also act as identity on $W_0$, meaning that $w_0$ would be in the center of $W_0$, that is a contradiction.
So, where has something gone wrong in my arguments ?
Edit: An example, as demanded in the comments.
I consider $G=\mathrm{GL}_n$ over an algebraically closed field $k$ with $\mathrm{char}(k)\not = 2$, and I assume that $n=2k+1$ is odd. I consider $T_0$ the usual maximal torus consisting of diagonal matrices, and $W_0$ is identified with $\mathfrak S_n$. I consider $w_0$ defined by the formula $w_0(i)=n+1-i$. As a representative in $N_G(T_0)$, I consider $n_0$ the matrix having antidiagonal filled with $1$'s.
Now, I put
$$g =
\begin{bmatrix}
1 & & & & & & 1\\
& \unicode{x22f1} & & & & \unicode{x22F0} & \\
& & 1 & & 1 & & \\
& & & 1 & & & \\
& & 1 & & -1 & & \\
& \unicode{x22F0} & & & & \unicode{x22f1} & \\
1 & & & & & & -1
\end{bmatrix}
\quad
g^{-1} =
\frac{1}{2}
\begin{bmatrix}
1 & & & & & & 1\\
& \unicode{x22f1} & & & & \unicode{x22F0} & \\
& & 1 & & 1 & & \\
& & & 2 & & & \\
& & 1 & & -1 & & \\
& \unicode{x22F0} & & & & \unicode{x22f1} & \\
1 & & & & & & -1
\end{bmatrix}
$$
This shape is possible because $n$ is odd. The matrix $g$ is invertible because of the assumption on the characteristic. Its inverse is given by the matrix on the right. I may now compute $g^{-1}\phi(g)$ which gives me the antidiagonal matrix having coefficients, from bottom left to top right, $k$ times $-1$ and $k+1$ times $1$. It is a generalized permutation matrix, so it lies in $N_G(T_0)$. In means that $T:=\,^gT_0$ is stable by $\phi$. Moreover, let us note that $\tilde{n_0}:=g^{-1}\phi(g)$ also is a representative of $w_0$, just as in the situation I considered above.
We may further compute $T$ explicitly. It is the subgroup (maximal torus) consisting of all cross-shaped invertible matrices (ie. with coefficients on the diagonals only) which are invariant under 180° rotation.
Edit2: Partial answer and thoughts.
I think that my mistake lies in "if $\phi$ acts like $\mathrm{id}$ on $W$ then it must act like $\mathrm{id}$ in $W_0$ as well".
Say, generally speaking, that $\phi:G\rightarrow G$ is an endomorphism of $G$ having $T_0$ and $T:=\,^gT_0$ as two maximal $\phi$-stable tori, so that $\tilde{n_0}:=g^{-1}\phi(g) \in N_G(T_0)$. Assume also that $\phi$ acts like $\mathrm{id}$ on $W$. It means precisely that $\forall \nu \in N_G(T), \;\nu^{-1}\phi(\nu)\in T$.
In terms of $T_0$, it means that $\forall \nu_0\in N_G(T_0),\; \nu_0^{-1}\tilde{n_0}\phi(\nu_0)\tilde{n_0}^{-1}\in T_0$. In other words, the composition of the conjugation by $\tilde{n_0}$ with $\phi$ induces the identity on $W_0$. Thus, $\phi$ acts like conjugation by (the image of) $\tilde{n_0}^{-1}$ on $W_0$, which may not be the identity. This is consistent with my situation above, where $\phi$ acts like conjugation by $w_0$.
The conclusion is that the action of an endomorphism on a Weyl group actually depends on the choice of the stable maximal torus ; that was unexpected for me.