Can all matrix Lie algebra automorphisms be expressed as
$A \rightarrow R^{-1} A R$ for some invertible $R$?
By this, I mean taking all linear transformations between a vector space and itself, and mapping each to another linear transformation (also between that same vector space and itself) such that the commutator is preserved.
You are asking whether for a matrix Lie algebra $\mathfrak{g}$, all automorphisms are inner automorphisms. The answer is no in general. E.g. for $n \ge 3$, the map
$$X \mapsto -X^{t}$$
(negative of the transpose) is an automorphism on the Lie algebras $\mathfrak{sl}_n(\Bbb C)$ and $\mathfrak{gl}_n(\Bbb C)$ which is not given by conjugation with a matrix. (Fun exercise: For $\mathfrak{sl}_2(\Bbb C)$, it is still an automorphism but an inner one, i.e. it is induced by conjugation with a certain matrix.)
In general, the automorphisms of simple (split) Lie algebras are pretty well understood. The existence of outer automorphisms can be read of the Dynkin diagram, and they exist in types $A_{n\ge 2}, D_{n \ge 3}$ and $E_6$. For a discussion see https://mathoverflow.net/q/14735/27465 and ch. VIII §5 of Bourbaki's Lie Groups and Algebras.
For other types of Lie algebras which might be represented as matrices (e.g. solvable ones, and even non-split semisimple ones over arbitrary fields), I have no idea, but I guess the answer is still "no" in general.