Let $G$ be a linear algebraic group over an algebraically closed field $k$, and $T$ a maximal torus of $G$.


Suppose we have two cocharacter $\mu, \mu' : \mathbb{G}\_m \to T$, which are conjugate under $G$ i.e. there exists $g \in G$ such that $\mu'(z) = g\mu(z)g^{-1}$.


**Question.** Can we always choose $g \in G$ such that $\mu'(z) = g\mu(z)g^{-1}$ and such that $g$ normalizes $T$?


If $G = \mathrm{GL}\_n$ then this is true: suppose that $T$ is the diagonal torus. Then a cocharacter of $T$ is just determined by its weights on the basis vectors of the natural representation of $G$ (i.e. integers $k\_i$ such that $\mu(z)e\_i = z^{k\_i} e\_i$).


Since $\mu'$, $\mu$ are $G$-conjugate, the weights of $\mu$ are a permutation of the weights of $\mu'$. We can choose an element of $G$ which induces this permutation on the basis vectors.


For general $G$, we can embed $G$ in $\mathrm{GL}\_n$ for some $G$, such that $T$ is the intersection of $G$ with the diagonal torus. The weights of $\mu$ on the basis vectors of the corresponding representation are a permutation of the weights of $\mu'$. The question is whether we can choose this permutation (since the weights need not be distinct, there may be several choices for the permutation) so that it is induced by $N\_G(T)$.


For my application, to show that the number of $G$-conjugates of $\mu$ in $T$ is finite, it is enough to know this for $\mathrm{GL}\_n$, but it would be nice to know if it is true in general.



Conjugate cocharacters in a maximal torus
Yes. That $\mu'(z) = g \mu(z) g^{-1}$ means that $g T g^{-1}$ centralizes the image of $\mu'$. Thus, $T$ and $g T g^{-1}$ are maximal tori in the centralizer
of the image of $\mu'$, so there exists $h$ in this centralizer such that $g T g^{-1} = h T h^{-1}$. Now replace $g$ by $h^{-1} g$. It doesn't seem that you need G to be semisimple or reductive for this, but I may be missing something here?


