Is an $\mathfrak{sl}_2$-triple determined up to Lie algebra automorphism by the adjoint representation?

Question

Let $\mathfrak{g}$ be a finite-dimensional complex semisimple Lie algebra, and let $\phi_1:\mathfrak{sl}_2(\mathbb{C})\rightarrow\mathfrak{g}$ and $\phi_{2}:\mathfrak{sl}_2(\mathbb{C})\rightarrow\mathfrak{g}$ be complex Lie algebra morphisms. By composing $\phi_1$ and $\phi_2$ with the adjoint representation of $\mathfrak{g}$, we obtain two representations of $\mathfrak{sl}_2(\mathbb{C})$ on $\mathfrak{g}$. My question is then the following.

Question: If these two $\mathfrak{sl}_2(\mathbb{C})$-representations are isomorphic, does it follow that $\phi_1$ and $\phi_2$ are related by an element of the Lie algebra automorphism group $\text{Aut}(\mathfrak{g})$?

To provide some context, suppose that $\mathfrak{g}=\mathfrak{so}_{4n}(\mathbb{C})$. If $\lambda$ is a partition of $4n$ having only even parts with each part appearing an even number of times, then $\lambda$ corresponds to exactly two distinct nilpotent orbits, $\mathcal{O}_1$ and $\mathcal{O}_2$, in $\mathfrak{so}_{4n}(\mathbb{C})$. Now, let $\phi_1:\mathfrak{sl}_2(\mathbb{C})\rightarrow\mathfrak{so}_{4n}(\mathbb{C})$ and $\phi_2:\mathfrak{sl}_2(\mathbb{C})\rightarrow\mathfrak{so}_{4n}(\mathbb{C})$ be Lie algebra maps satisfying $\phi_1(e)\in\mathcal{O}_1$ and $\phi_2(e)\in\mathcal{O}_2$. As discussed above, each map gives a representation of $\mathfrak{sl}_2(\mathbb{C})$ on $\mathfrak{so}_{4n}(\mathbb{C})$. These two representations are actually isomorphic. While $\phi_1$ and $\phi_2$ cannot be related by an inner automorphism of $\mathfrak{so}_{4n}(\mathbb{C})$ (as $\mathcal{O}_1\neq\mathcal{O}_2$), they are nevertheless related by a Lie algebra automorphism.

Answer 1

REVISED VERSION: On further reflection, I think the answer to your question is always "yes" (if $\mathfrak{g}$ is simple, to avoid complications of the type Dave indicates). At first I was confused by the example discussed in the question, but I think the main issue is how the automorphism group interacts with Dynkin diagrams of nilpotent orbits.

In the classical Dynkin-Kostant treatment of nilpotent orbits in a simple Lie algebra $\mathfrak{g}$, the basic strategy is to embed a (nonzero) nilpotent element in a copy of $\mathfrak{sl}_2(\mathbb{C})$. Such an embedding isn't unique, but there is a resulting bijection between conjugacy classes of nilpotents (under the adjoint group of $\mathfrak{g}$) and conjugacy classes of such subalgebras. In turn, the adjoint action of the subalgebra leads to a decorated Dynkin diagram (with vertices labelled $0, 1, 2$) which determines uniquely the given nilpotent orbit. (Some references to textbooks by Carter and Collingwood-McGovern which include Dynkin diagrams are given in my old notes here.)

In your type $D$ example, there are typically pairs of orbits interchanged by an outer automorphism of $\mathfrak{g}$ that comes from a graph automorphism. These orbits have many common properties, and corresponding copies of $\mathfrak{sl}_2(\mathbb{C})$ do act equivalently on $\mathfrak{g}$ even though they lead to distinct Dynkin diagrams. The key fact is that these Dynkin diagrams just involve a permutation of labels induced by the outer automorphism. Only in such limited cases can two subalgebras of type $\mathfrak{sl}_2(\mathbb{C})$ act equivalently in the adjoint representation of $\mathfrak{g}$: this is clear from the determination of Dynkin diagrams for each $\mathfrak{g}$.

[Concerning terminology, it's fairly conventional to call two representations equivalent but the associated modules isomorphic. The choice of either representation or module language is usually optional, at least in finite dimensional cases.]

Answer 2

No, here is a counterexample.

Let $\mathfrak{g}_1 = \mathfrak{sl}_2(\mathbb{C}) \oplus \mathfrak{sl}_2(\mathbb{C}) \oplus \mathfrak{sl}_2(\mathbb{C})$, and let $\phi_1 \colon \mathfrak{sl}_2(\mathbb{C}) \to \mathfrak{g}_1$ be the diagonal embedding, so $\phi(x) = (x,x,x)$. Then the representation obtained by composing $\phi_1$ with the adjoint representation is the direct sum of three $3$-dimensional irreducible modules.

Let $\mathfrak{g}_2 = \mathfrak{so}_5(\mathbb{C})$, and let $\phi_2$ be the embedding onto $\langle \mathfrak{u}_\alpha, \mathfrak{u}_{-\alpha} \rangle$, where $\alpha$ is a short root. Then the representation obtained by composing $\phi_2$ with the adjoint representation is the direct sum of a trivial representation and precisely three $3$-dimensional irreducible modules.

Now, if we let $\mathfrak{g} = \mathfrak{g}_1 \oplus \mathfrak{g}_2$, then we can think of $\phi_1$ and $\phi_2$ as homomorphisms into $\mathfrak{g}$ and, from the above, we see that their compositions with the adjoint representation are isomorphic. However, the image of $\phi_2$ is contained in the unique ideal of $\mathfrak{g}$ that is isomorphic to $\mathfrak{so}_5(\mathbb{C})$, but the image of $\phi_1$ is not contained in this ideal. So no element of $\mathrm{Aut}(\mathfrak{g})$ can take $\phi_1$ to $\phi_2$.