Are Lie algebra representations $\rho$ and $\rho\circ F$ isomorphic? Let $F: \mathfrak g \to \mathfrak g$ be a Lie algebra automorphism, and let $\rho: \mathfrak g \to GL(V)$ be a representation.
Question: Are $\rho$ and $\rho \circ F$ isomorphic?
Motivation: This question is motivated from the physics literature. Physicists often deal with 2nd quantization formalism (where operators can be thought of as elements of a Lie algebra), and they sometimes employ transformations on these operators that preserve the commutation relations. Then, they (maybe naively) use the transformed operators. When I carefully examine these arguments, there is a gap which is precisely stated by my question.
 A: In general, these representations are not equivalent.
Example: Let $\mathfrak g := \mathfrak{sl}_3(\mathbb C)$ and $F: X \mapsto -X^{tr}$. Let $\rho: \mathfrak{sl}_3(\mathbb C) \subset \mathfrak{gl}_3(\mathbb C)$ the standard representation. Then it is an exercise to check that there is no $A \in M_3(\mathbb C)$ such that $F(X) = AXA^{-1}$ for all $X\in \mathfrak g$, i.e. the representations $\rho$ and $\rho \circ F$ are not equivalent.

However, note the following which might redeem the approach: An isomorphism of Lie algebras $F: \mathfrak g_1 \stackrel{\simeq}\rightarrow \mathfrak g_2$ does induce an equivalence of categories between (say, finite-dimensional complex) representations of $\mathfrak g_1$ and (finite-dimensional complex) representations of $\mathfrak g_2$ via
$$ Rep (\mathfrak g_2) \rightarrow Rep(\mathfrak g_1)$$
$$ (\rho, V) \mapsto (\rho \circ F, V).$$
In particular, any automorphism $F: \mathfrak g \rightarrow \mathfrak g$ induces an auto-equivalence of its category of representations, however as the above example shows, this auto-equivalence does not need to be isomorphic to the identity. However, if our interest is to investigate the general structure of that category, i.e. all (fin.-dim., complex) representations, this is kind of harmless. In a way, we only "shuffle them around".
[Actually, in the above example, which generalises to all $\mathfrak{sl}_{n \ge 3}$, the involution $F$, being "the" outer automorphism of the Lie algebra, on first sight seems to induce (up to ...) the duality functor on the category, i.e.: $$\rho \circ F \simeq \rho^* $$
However it is not actually the duality functor, because that one also flips arrows around (is contravariant), whereas the one defined as above via $F$ is covariant: They only do the same on objects. Now this estranged me at first, but I think is redeemed by the category also being semisimple (and duals commuting with direct sums), i.e. basically every arrow can be reversed.]
