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.



Is an $\mathfrak{sl}_2$-triple determined up to Lie algebra automorphism by the adjoint representation?
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](http://people.math.umass.edu/~jeh/pub/D_4.pdf).)


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.]


