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REVISED VERSION: On further reflection, I think the answer to your question is always "yes". At first I accepted too uncritically the example discussed in the question, where there is an unsupported statement which strikes me now as incorrect: "These two representations are actually isomorphic." This seems to be the origin of the apparent contradictions which led to the question.

In the classical Dynkin-Kostant treatment of nilpotent orbits in $\mathfrak{g}$, 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. While these orbits do have many common properties, corresponding copies of $\mathfrak{sl}_2(\mathbb{C})$ can't act equivalently on $\mathfrak{g}$ since they lead to distinct Dynkin diagrams.

[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.]

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.]

REVISED VERSION: On further reflection, I think the answer to your question is always "yes". At first I accepted too uncritically the example discussed in the question, where there is an unsupported statement which strikes me now as incorrect: "These two representations are actually isomorphic." This seems to be the origin of the apparent contradictions which led to the question.

In the classical Dynkin-Kostant treatment of nilpotent orbits in $\mathfrak{g}$, 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 come from a graph automorphism. While these orbits do have many common properties, corresponding copies of $\mathfrak{sl}_2(\mathbb{C})$ can't act equivalently on $\mathfrak{g}$ since they lead to distinct Dynkin diagrams.

[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.]

REVISED VERSION: On further reflection, I think the answer to your question is always "yes". At first I accepted too uncritically the example discussed in the question, where there is an unsupported statement which strikes me now as incorrect: "These two representations are actually isomorphic." This seems to be the origin of the apparent contradictions which led to the question.

In the classical Dynkin-Kostant treatment of nilpotent orbits in $\mathfrak{g}$, 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. While these orbits do have many common properties, corresponding copies of $\mathfrak{sl}_2(\mathbb{C})$ can't act equivalently on $\mathfrak{g}$ since they lead to distinct Dynkin diagrams.

[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.]

[EDITED] It's probably better to assume that $\mathfrak{g}$ is simple, to avoid complications, and that the representations are faithful (hence equivalent).

Maybe I'm overlooking something subtle, but I think the answer is no if there are two root lengths involved: then the root vectors targeted by $\phi_1, \phi_2$ can't be related by a Lie algebra automorphism.

In the other direction, I think the answer is yes if all roots have the same length (as in your example) and the two roots involved here are in the same orbit under inner automorphisms. If they lie in different orbits (as in your example), it will depend on whether there exist outer automorphisms which interchange these orbits. That seems to require case-by-case classification.

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

REVISED VERSION: On further reflection, I think the answer to your question is always "yes". At first I accepted too uncritically the example discussed in the question, where there is an unsupported statement which strikes me now as incorrect: "These two representations are actually isomorphic." This seems to be the origin of the apparent contradictions which led to the question.

In the classical Dynkin-Kostant treatment of nilpotent orbits in $\mathfrak{g}$, 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 come from a graph automorphism. While these orbits do have many common properties, corresponding copies of $\mathfrak{sl}_2(\mathbb{C})$ can't act equivalently on $\mathfrak{g}$ since they lead to distinct Dynkin diagrams.

[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.]