Let $G$ be a connected, reductive linear algebraic group whose semisimple rank is $1$. Then $H := (G,G)$ is a connected semisimple group of rank one. Let $T_1$ be a maximal torus of $H$, and let $T$ be a maximal torus of $G$ containing $T_1$. Is it true that $$N_H(T_1) \subseteq N_G(T)$$ and that $$T_1 = N_H(T_1) \cap T?$$ If so, the inclusion $N_H(T_1) \subseteq N_G(T)$ would induce an isomorphism of Weyl groups $W(G,T) \cong W(H,T_1)$.
In the single example I have worked, this is the case: Let $G = \textrm{GL}_2$, and $H = (G,G) = \textrm{SL}_2$. Take $$T_1 = \{ \begin{pmatrix} x & 0 \\ 0& x^{-1} \end{pmatrix} : x \in k^{\ast} \}$$ and $$T = \{ \begin{pmatrix} x & 0 \\ 0& y \end{pmatrix} : x,y \in k^{\ast} \}$$ Set $n = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$. And note that $n \in N_H(T_1) \setminus T_1$. Since we know that $N_H(T_1)/T_1$ has order two, it follows that $$N_H(T_1) = T_1 \cup nT_1 $$ So in this case it's easy to check that $N_H(T_1) \subseteq N_G(T)$ and $T_1 = N_H(T_1) \cap T$.
I think it's true? Not sure if this is right. If $R(G)$ is the radical, then the product map $R(G) \times T_1 \rightarrow T$ is a morphism of algebraic groups with kernel $R(G) \cap T_1$, which is finite. The image is $R(G).T_1$, which is a closed, connected subgroup of $T$. The dimension of the image is $$\textrm{Dim } R(G) \times T_1 - \textrm{Dim } R(G) \cap T_1 = \textrm{Dim } R(G) + \textrm{Dim } T_1 - 0 = \textrm{Dim } T$$ so $R(G)$ and $T_1$ generate $T$. Any element of $N_G(T_1)$ normalizes $R(G)$ and $T_1$, so it must normalize $T$.
So this shows that $N_H(T_1) \subseteq N_G(T_1) \subseteq N_G(T)$.
Now $N_H(T_1) \cap T$ is a subgroup of $N_H(T_1)$ containing $T_1$. The quotient $N_H(T_1)/T_1$ is $\mathbb{Z}/2\mathbb{Z}$, so either $N_H(T_1) \cap T = T_1$ or $N_H(T_1) \cap T = N_H(T_1)$.
This last case can't happen. If $N_H(T_1) \cap T = N_H(T_1)$, then $N_H(T_1) \subseteq T$. Let $n \in N_H(T_1)$ but not in $T_1$. Then we know that $n$ has the effect $ntn^{-1} = t^{-1}$ for all $t \in T_1$ (classification of automorphisms of the affine line). But $n \in T$, so $ntn^{-1} = t$ for all $t \in T_1$, contradiction.