Let $B$ be a Borel subgroup containing $T$. As $F(B)$ and $B$ are both Borel subgroups containing $T$ there exists an element $n \in N_G(T)$ such that ${}^nF(B) = B$. Thus the Frobenius endomorphism $F' : G \to G$ defined by $F'(g) = nF(g)n^{-1}$ induces an automorphism $F' : W \to W$, where $W = N_G(T)/T$, and this stabilises the Coxeter generators $\mathbb{S} \subseteq W$ determined by $B$. Thus $F'$ is a length preserving automorphism so must fix the longest element (by uniqueness). Hence, $F$ fixes the longest element if and only if $\bar{n} \in C_W(w_0)$, where $\bar{n} \in W$ is the image of $n$.

One can ask how unique the element $n$ is. Assume $m \in N_G(T)$ also satisfies ${}^mF(B) = B$ then $mn^{-1} \in N_G(B) = B$ so $mn^{-1} \in N_B(T) = T$ thus $\bar{n} = \bar{m}$. Hence the condition that $F$ fixes the longest element does not depend upon the choice of element $n$.

Let's see how to compute this in practice. Choose an $F$-stable maximal torus and Borel subgroup $T_0 \leqslant B_0 \leqslant G$. There then exists an element $x \in G$ such that $(T,B) = ({}^xT_0,{}^xB_0)$. Note, by the above argument that such an element $x$ is unique up to right multiplication by elements of $T_0$. Let $(W_0,\mathbb{S}_0)$ and $(W,\mathbb{S})$ be the Coxeter system defined with respect to $(T_0,B_0)$ and $(T,B)$ respectively. Then conjugation by $x$ induces an isomorphism $W_0 \to W$ mapping $\mathbb{S}_0$ onto $\mathbb{S}$. Hence, if $w_0 \in W_0$ is the longest element then ${}^xw_0 \in W$ is the longest element.

Assume $x^{-1}F(x) = n \in N_G(T_0)$ then we have $$F(B) = F({}^xB_0) = {}^{xn}B_0 = {}^{xnx^{-1}}B.$$ Hence $xn^{-1}x^{-1} \in N_G(T) = {}^xN_G(T_0)$ is an element as above. We have $x\bar{n}^{-1}x^{-1} \in C_W({}^xw_0) = {}^xC_{W_0}(w_0)$ if and only if $\bar{n} \in C_{W_0}(w_0)$. Thus one can easily construct examples where the longest element is not fixed.

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