Adjoint action of $\mathrm{SL}_2$ on a Cartan subalgebra Let $G = \mathrm{SL}_2(\mathbb{R})$ and $\mathfrak{g}$ its Lie algebra. We define the adjoint action of $G$ on the Lie algebra $\mathfrak{g}$ like the map $\mathrm{Ad} \colon G \times \mathfrak{g} \to \mathfrak{g}$ with $\mathrm{Ad}(g)(X) = g X g^{-1}$.
We define the Lie bracket like the map $\mathrm{ad} \colon \mathfrak{g} \times \mathfrak{g} \to \mathfrak{g}$ given by $\mathrm{ad}(X,Y) = [X,Y] = XY-YX$.
I know that we have relations like $\mathrm{ad}(h,f) = [h,f] = -2f$ when $f \in \mathfrak{g}$ is lower triangular and $h$ in Cartan subalgebra.
Can we deduce relations like that when we consider the action of the Lie group on a Cartan subalgebra? I.e. the value of $\mathrm{Ad}(F)(h) = FhF^{-1}$ for $F$ in the unipotent radical and $h$ in a Cartan subalgebra?
 A: You don't specify the unipotent radical of what but I assume you mean of the subgroup of lower triangular matrices since $f$ is lower triangular (indeed you want it strictly lower triangular). The nilpotent radical $\mathfrak{n}$ of the lower triangular subalgebra $\mathfrak{b}$ is isomorphic to the unipotent radical in the lower triangular subgroup via its exponential map since $\mathfrak{n}$ is nilpotent. In other words, $F = \exp(f)$ for some $f\in\mathfrak{n}$. So you are trying to compute $\operatorname{Ad}(\exp(f))(h)$ but this is the same as
$$\exp(\operatorname{ad}(f))(h) = h + [f,h]+\frac{1}{2}[f,[f,h]] + \dots$$
Now $[f,h]= 2f$ and $[f,[f,h]]=0$ and indeed the rest of the terms vanish. So $\operatorname{Ad}(F)(h) = h + 2f$.
Note this will of course extend to other $\mathfrak{sl}_n$ although of course it will get a little bit more complicated as we would need to account for different possibilities. For example, at what stage is $\operatorname{ad}^n(f)=0$, is $f$ a root vector and for which root, is $h$ a coroot and so on.
