Proposition 5

— Assume that $\mathfrak g$ is splittable. The group $\operatorname{Aut}_0(\mathfrak g)$ operates simply-transitively on the set of framings of $\mathfrak g$.

Let $e_1=(\mathfrak g,\mathfrak h_1,B_1,(X^1_\alpha)_{\alpha\in B_1})$, $e_2=(\mathfrak g,\mathfrak h_2,B_2,(X^2_\alpha)_{\alpha\in B_2})$ be two framings of $\mathfrak g$. There exists at least one element of $\operatorname{Aut}_0(\mathfrak g)$ that transforms $e_1$ into $e_2$ (Prop. 1 and Prop. 4). Let $\overline{k}$ be an algebraic closure of $k$. There exists an element of $\operatorname{Aut}_e(\mathfrak g\otimes_k\overline{k})$ that transforms $\mathfrak h_1\otimes_k\overline{k}$ into $\mathfrak h_2\otimes_k\overline{k}$ (Chap. VII, §3, no. 2, Th. 1). Hence, by Prop. 4 and Cor. 2 of Prop. 2, there exists an element $\varphi$ of $\operatorname{Aut}_e(\mathfrak g\otimes_k\overline{k})$ that transforms the framing $(\mathfrak g\otimes_k\overline{k},\mathfrak h_1\otimes_k\overline{k},B_1,(X^1_\alpha)_{\alpha\in B_1})$ of $\mathfrak g\otimes_k\overline{k}$ into the framing $(\mathfrak g\otimes_k\overline{k},\mathfrak h_2\otimes_k\overline{k},B_2,(X^2_\alpha)_{\alpha\in B_2})$. Since $\mathfrak h_1$ and the $X^1_\alpha$ (resp. $\mathfrak h_2$ and the $X^2_\alpha$) generate $\mathfrak g_1$ (resp. $\mathfrak g_2$), we have $\varphi(\mathfrak g_1)=\mathfrak g_2$, so $\varphi$ is of the form $\psi\otimes 1$ where $\psi\in\operatorname{Aut}_0(\mathfrak g)$, and $\psi$ transforms $e_1$ into $e_2$.