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With the $G(k)$-conjugacy settled, consider the group $W := N_ {G(k)}(S)/Z_ {G(k)}(S)$. I claim that this is always a finite group, even with $S$ replaced by any $k$-torus in $G$. This can be proved in several ways. Here is one. The functorial normalizer and centralizer of $S$ are defined in Definition A.1.9 of "Pseudo-reductive groups", immediately after which it is proved that they are represented by closed $k$-subgroup schemes $N_G(S)$ and $Z_G(S)$, with $Z_G(S)$ normal in $N_G(S)$ from the definitions. The quotient $W(G,S) := N_G(S)/Z_G(S)$ then makes sense as a finite type $k$-group, and (as is also shown in SGA3 by the same method) the proof of Lemma A.2.9 shows that $W(G,S)$ is 'etale and hence $k$-finite (the smoothness of $G$ which is assumed there isn't relevant to that part of the proof). Thus, $W(G,S)(k)$ is finite, and $W$ is a subgroup of this (since $N_ {G(k)}(S) = N_G(S)(k)$ and $Z_ {G(k)}(S) = Z_G(S)(k)$).

Theorem: Let ${\rm{X}}_ {\ast}(G) = {\rm{Hom}}_ k(\mathbf{G}_ m, G)$, and likewise for ${\rm{X}}_ {\ast}(S)$. Under the natural $G(k)$-action on ${\rm{X}}_ {\ast}(G)$ and the natural $W$-action on ${\rm{X}}_ {\ast}(S)$, the natural map $$W\backslash {\rm{X}}_ {\ast}(S) \rightarrow G(k) \backslash {\rm{X}}_ {\ast}(G)$$ is bijective. In particular, $G(k)$-orbits on ${\rm{X}}_ {\ast}(G)$ have finite non-empty intersection with ${\rm{X}}_ {\ast}(S)$.

Proof: Letting $G' \subseteq G$ denote the Galois descent of the Zariski closure of $G(k_s)$ in $G_ {k_s}$, so $G'$ is a smooth closed $k$-subgroup of $G$ (see Lemma C.4.1 once again), clearly ${\rm{X}}_ {\ast}(G') = {\rm{X}}_ {\ast}(G)$, $G'(k) = G(k)$, and $S$ is contained in $G'$. Thus, we can replace $G$ with $G'$ without affecting what is to be proved, and so we can assume $G$ is smooth. We now adapt the argument from the proof of Lemma C.3.5 (which treats a variant for pseudo-reductive $k$-groups and maximal $k$-tori). Any $k$-homomorphism $\mathbf{G}_ m \rightarrow G$ has image which is a $k$-split $k$-torus, so it lands in a maximal $k$-split $k$-torus. But all such $k$-tori are $G(k)$-conjugate to our friend $S$, as explained above (even without smoothness), so we get the surjectivity.

With the $G(k)$-conjugacy settled, consider the group $W := N_ {G(k)}(S)/Z_ {G(k)}(S)$. I claim that this is always a finite group, even with $S$ replaced by any $k$-torus in $G$ and without smoothness hypotheses on $G$ (a red herring, by the trick in the preceding paragraph). This can be proved in several ways. Here is one. The functorial normalizer and centralizer of $S$ are defined in Definition A.1.9 of "Pseudo-reductive groups", immediately after which it is proved that they are represented by closed $k$-subgroup schemes $N_G(S)$ and $Z_G(S)$, with $Z_G(S)$ normal in $N_G(S)$ from the definitions. The quotient $W(G,S) := N_G(S)/Z_G(S)$ then makes sense as a finite type $k$-group, and (as is also shown in SGA3 by the same method) the proof of Lemma A.2.9 shows that $W(G,S)$ is 'etale and hence $k$-finite (the smoothness of $G$ which is assumed there isn't relevant to that part of the proof). Thus, $W(G,S)(k)$ is finite, and $W$ is a subgroup of this (since $N_ {G(k)}(S) = N_G(S)(k)$ and $Z_ {G(k)}(S) = Z_G(S)(k)$).

Theorem: Let $G$ be an affine group of finite type over a field $k$, and $S$ a maximal $k$-split $k$-torus. Define $W$ as above, let ${\rm{X}}_ {\ast}(G) = {\rm{Hom}}_ k(\mathbf{G}_ m, G)$, and likewise for ${\rm{X}}_ {\ast}(S)$.

Under the natural $G(k)$-action on ${\rm{X}}_ {\ast}(G)$ and the natural $W$-action on ${\rm{X}}_ {\ast}(S)$, the natural map $$W\backslash {\rm{X}}_ {\ast}(S) \rightarrow G(k) \backslash {\rm{X}}_ {\ast}(G)$$ is bijective. In particular, $G(k)$-orbits on ${\rm{X}}_ {\ast}(G)$ have finite non-empty intersection with ${\rm{X}}_ {\ast}(S)$.

Proof: Letting $G' \subseteq G$ denote the Galois descent of the Zariski closure of $G(k_s)$ in $G_ {k_s}$, so $G'$ is a smooth closed $k$-subgroup of $G$ (see Lemma C.4.1 once again), clearly ${\rm{X}}_ {\ast}(G') = {\rm{X}}_ {\ast}(G)$, $G'(k) = G(k)$, and $S$ is contained in $G'$. Thus, we can replace $G$ with $G'$ without affecting what is to be proved, and so we can assume $G$ is smooth (but maybe not connected).
We now adapt the argument from the proof of Lemma C.3.5 (which treats a variant for pseudo-reductive $k$-groups and maximal $k$-tori). Any $k$-homomorphism $\mathbf{G}_ m \rightarrow G$ has image which is a $k$-split $k$-torus, so it lands in a maximal $k$-split $k$-torus. But all such $k$-tori are $G(k)$-conjugate to our friend $S$, as explained above (even without smoothness), so we get the surjectivity.

In fact something better is true (properly formulated!) over any field $k$, using $k$-rational conjugacy and maximal $k$-split $k$-tori. I will give a precise statement and proof below, with $G$ any affine $k$-group of finite type (no reductivity or smoothness hypotheses, though lack of smoothness is a red herring, as will be explained below; lack of reductivity is more serious with general $k$). In the end it will come down to the same idea as in the argument which is given by an$\overline{\rm{a}}$dimadhy$\overline{\rm{a}}$nta, but we need preparations to make it work over any field and for any $G$ (even with smoothness restrictions, there is work to do). To stick with a single reference, below I refer to things in "Pseudo-reductive groups" (no surprise there).

In fact something better is true (properly formulated!) over any field $k$, using $k$-rational conjugacy and maximal $k$-split $k$-tori. I will give a precise statement and proof below, with $G$ any affine $k$-group of finite type (no reductivity or smoothness hypotheses, though lack of smoothness is a red herring, as will be explained below; lack of reductivity is more serious with general $k$). In the end it will come down to the same idea as in the argument which is given by an$\overline{\rm{a}}$dimadhy$\overline{\rm{a}}$nta, but we need preparations to make it work over any field and for any $G$ (even with smoothness restrictions, there is work to do). To stick with a single reference, below I refer to things in "Pseudo-reductive groups".

With the $G(k)$-conjugacy settled, consider the group $W := N_ {G(k)}(S)/Z_ {G(k)}(S)$. I claim that this is always a finite group, even with $S$ replaced by any $k$-torus in $G$. (This finiteness will not be used, so a reader who doesn't care can skip its proof.) This can be proved in several ways. Here is one. The functorial normalizer and centralizer of $S$ are defined in Definition A.1.9 of "Pseudo-reductive groups", immediately after which it is proved that they are represented by closed $k$-subgroup schemes $N_G(S)$ and $Z_G(S)$, with $Z_G(S)$ normal in $N_G(S)$ from the definitions. The quotient $W(G,S) := N_G(S)/Z_G(S)$ then makes sense as a finite type $k$-group, and (as is also shown in SGA3 by the same method) the proof of Lemma A.2.9 shows that $W(G,S)$ is 'etale and hence $k$-finite (the smoothness of $G$ which is assumed there isn't relevant to that part of the proof). Thus, $W(G,S)(k)$ is finite, and $W$ is a subgroup of this (since $N_ {G(k)}(S) = N_G(S)(k)$ and $Z_ {G(k)}(S) = Z_G(S)(k)$).

Theorem: Let ${\rm{X}}_ {\ast}(G) = {\rm{Hom}}_ k(\mathbf{G}_ m, G)$, and likewise for ${\rm{X}}_ {\ast}(S)$. Under the natural $G(k)$-action on ${\rm{X}}_ {\ast}(G)$ and the natural $W$-action on ${\rm{X}}_ {\ast}(S)$, the natural map $$W\backslash {\rm{X}}_ {\ast}(S) \rightarrow G(k) \backslash {\rm{X}}_ {\ast}(G)$$ is bijective.

With the $G(k)$-conjugacy settled, consider the group $W := N_ {G(k)}(S)/Z_ {G(k)}(S)$. I claim that this is always a finite group, even with $S$ replaced by any $k$-torus in $G$.  This can be proved in several ways. Here is one. The functorial normalizer and centralizer of $S$ are defined in Definition A.1.9 of "Pseudo-reductive groups", immediately after which it is proved that they are represented by closed $k$-subgroup schemes $N_G(S)$ and $Z_G(S)$, with $Z_G(S)$ normal in $N_G(S)$ from the definitions. The quotient $W(G,S) := N_G(S)/Z_G(S)$ then makes sense as a finite type $k$-group, and (as is also shown in SGA3 by the same method) the proof of Lemma A.2.9 shows that $W(G,S)$ is 'etale and hence $k$-finite (the smoothness of $G$ which is assumed there isn't relevant to that part of the proof). Thus, $W(G,S)(k)$ is finite, and $W$ is a subgroup of this (since $N_ {G(k)}(S) = N_G(S)(k)$ and $Z_ {G(k)}(S) = Z_G(S)(k)$).

Theorem: Let ${\rm{X}}_ {\ast}(G) = {\rm{Hom}}_ k(\mathbf{G}_ m, G)$, and likewise for ${\rm{X}}_ {\ast}(S)$. Under the natural $G(k)$-action on ${\rm{X}}_ {\ast}(G)$ and the natural $W$-action on ${\rm{X}}_ {\ast}(S)$, the natural map $$W\backslash {\rm{X}}_ {\ast}(S) \rightarrow G(k) \backslash {\rm{X}}_ {\ast}(G)$$ is bijective. In particular, $G(k)$-orbits on ${\rm{X}}_ {\ast}(G)$ have finite non-empty intersection with ${\rm{X}}_ {\ast}(S)$.