Question

The uniqueness can break down very badly in positive characteristic. Supose $G=SL_p$ where $p$ is the characteristic of the base field. Take a regular nilpotent element $e$ in $\mathfrak{g}=\mathfrak{sl}_p$. Then there is a nilpotent element $f\in\mathfrak{g}$ such that $e$, $f$ and $h=[e,f]$ form an $\mathfrak{sl}_2$-triple with the property that $h^p=h$. Note that the identity matrix $I$ is in $\mathfrak{g}$. It is easy to see that there is an $f_0\in\mathfrak{g}$ such that $[e,f_0]=I$ (many lecturers find this fact useful when explaining that Lie's theorem can fail in characteristic $p$). Let $\lambda$ be a scalar such that $\lambda^p\ne \lambda$. Since $h$ commutes with $I$ and $ad\ h$ is semisimple, we may assume further that $[h,f_0]=-2f_0$. Then $(e,h+\lambda I, f+\lambda f_0)$ is another $\mathfrak{sl}_2$-triple containing $e$. If the spans $\mathfrak{s}_1$ and $\mathfrak{s}_2$ of the triples are conjugate under $G$, then restricting the $p$-dimensional vector representaion of $\mathfrak{sl}_p$ to $\mathfrak{s}_1$ and $\mathfrak{s}_2$ we would get equivalent representations of $\mathfrak{sl}_2$. However, the representation we get from $\mathfrak{s}_1$ is restricted whereas the one we get from $\mathfrak{s}_2$ is not. So the triples are not conjugate under $C_G(e)$. There are similar examples for all One can replicate this example inside any Lie algebra of a reductive group $\mathfrak{sl}_n$ with \widetilde{G}$ whch contains $p|n$.G$ as a closed subgroup.