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If your representation $V$ is irreducible, it carries a unique up to scalars non-degenerate "contravariant form", with the property that $E$ and $F$ are adjoint. If you have chosen a self-dual basis (so the form looks like dot product) then the matrix form of $F$ is therefore the transpose of $E$. The matrix $H$ is determined by $H=[E,F]$.

In the general case there are of course many "contravariant forms" on a reducible finite dimensional representation $V$, but perhaps there is a natural choice in whatever situation you are interested in.

On the other hand, if your question is, given a nilpotent matrix $E$, is there a unique way to extend it to an $sl_2$-representation, the answer is no, but almost. See paragraph 11 of chapter VIII of Bourbaki's "Lie groups and Lie algebras" on $sl_2$-triplets. The bottom line is that if you want to pin down matrix forms of $F$ and $H$ just from knowledge of the matrix of $E$ you need a bit more structure (b/c you can conjugate $H$ and $F$ by anything in the centralizer of $E$).

Edited in light of clarifications made by OP:

Given a nilpotent matrix $E$ acting on a finite dimensional vector space $V$, it is always possible to extend it to a representation of $sl_2$ in such a way that it represents $e$. The extension is almost never unique: conjugating the representing matrices $F$ and $H$ by anything in the centralizer of $E$ gives a new extension.

The existence statement is the Jacobson-Morozov lemma (part of whose standard proof is reproduced in another answer) applied to the semisimple Lie algebra $sl(V)$. See Proposition 2 of section 2 of paragraph 11 of Bourbaki's "Lie Groups and Lie Algebras", Chapter VIII (see the Corollary following the Proposition for the extent to which uniqueness is true: basically, up to conjugacy).

On the other hand, if you have some additional rigid structure, there might be a unique extension. For instance, if you know a contravariant form and have an orthonormal basis at your disposal then $F$ is the transpose of $E$ (written in terms of the given orthonormal basis) and $H$ is determined by $H=[E,F]$.

If your representation $V$ is irreducible, it carries a unique up to scalars non-degenerate "contravariant form", with the property that $E$ and $F$ are adjoint. If you have chosen a self-dual basis (so the form looks like dot product) then the matrix form of $F$ is therefore the transpose of $E$. The matrix $H$ is determined by $H=[E,F]$.

In the general case there are of course many "contravariant forms" on a reducible finite dimensional representation $V$, but perhaps there is a natural choice in whatever situation you are interested in.

On the other hand, if your question is, given a nilpotent matrix $E$, is there a unique way to extend it to an $sl_2$-representation, the answer is no, but almost. See paragraph 11 of chapter VIII of Bourbaki's "Lie groups and Lie algebras" on $sl_2$-triplets. The bottom line is that if you want to pin down matrix forms of $F$ and $H$ just from knowledge of the matrix of $E$ you need a bit more structure.

If your representation $V$ is irreducible, it carries a unique up to scalars non-degenerate "contravariant form", with the property that $E$ and $F$ are adjoint. If you have chosen a self-dual basis (so the form looks like dot product) then the matrix form of $F$ is therefore the transpose of $E$. The matrix $H$ is determined by $H=[E,F]$.

In the general case there are of course many "contravariant forms" on a reducible finite dimensional representation $V$, but perhaps there is a natural choice in whatever situation you are interested in.

On the other hand, if your question is, given a nilpotent matrix $E$, is there a unique way to extend it to an $sl_2$-representation, the answer is no, but almost. See paragraph 11 of chapter VIII of Bourbaki's "Lie groups and Lie algebras" on $sl_2$-triplets. The bottom line is that if you want to pin down matrix forms of $F$ and $H$ just from knowledge of the matrix of $E$ you need a bit more structure (b/c you can conjugate $H$ and $F$ by anything in the centralizer of $E$).

If your representation $V$ is irreducible, it carries a unique up to scalars non-degenerate "contravariant form", with the property that $E$ and $F$ are adjoint. If you have chosen a self-dual basis (so the form looks like dot product) then the matrix form of $F$ is therefore the transpose of $E$. The matrix $H$ is determined by $H=[E,F]$.

In the general case there are of course many "contravariant forms" on a reducible finite dimensional representation $V$, but perhaps there is a natural choice in whatever situation you are interested in.

If your representation $V$ is irreducible, it carries a unique up to scalars non-degenerate "contravariant form", with the property that $E$ and $F$ are adjoint. If you have chosen a self-dual basis (so the form looks like dot product) then the matrix form of $F$ is therefore the transpose of $E$. The matrix $H$ is determined by $H=[E,F]$.

In the general case there are of course many "contravariant forms" on a reducible finite dimensional representation $V$, but perhaps there is a natural choice in whatever situation you are interested in.

On the other hand, if your question is, given a nilpotent matrix $E$, is there a unique way to extend it to an $sl_2$-representation, the answer is no, but almost. See paragraph 11 of chapter VIII of Bourbaki's "Lie groups and Lie algebras" on $sl_2$-triplets. The bottom line is that if you want to pin down matrix forms of $F$ and $H$ just from knowledge of the matrix of $E$ you need a bit more structure.