Given a Lie algebra $\mathfrak{g}$ and an ad-nipotent element $x$ of $\mathfrak{g}$, it can be shown that $\exp(\operatorname{ad} x)$ is a Lie algebra automorphism of $\mathfrak{g}$. Its inverse can be computed to be $\exp(-\operatorname{ad} x)$. Let us call an automorphism of this form a "pure inner automorphism". Note that the identity on $\mathfrak{g}$ writes as a pure inner automorphism, $\exp(\operatorname{ad}0)=1_\mathfrak{g}$. My question is: is the composition of two pure inner automorphism a pure inner automorphism? I was thinking that perhaps the Baker–Campbell–Hausdorff formula could generalize to this situation. If given ad-nilpotent elements $x,y\in\mathfrak{g}$ one were to look for an ad-nipotent element $z$ that solved the equation $$\exp(\operatorname{ad} x)\exp(\operatorname{ad} y)=\exp(\operatorname{ad} z),$$ I guess the sum of the iterated Lie brackets from the BCH should turn out to be finite, as $x$ and $y$ are ad-nilpotent, and maybe $z$ could be computed that way, using the fact that $\operatorname{ad}$ is a Lie algebra homomorphism (and assuming that "the bracket of ad-nilpotents elements is ad-nilpotent" is true, which I don't know to be that way).
Typically, in the literature one sees that they define the "subgroup of inner automorphisms," $\operatorname{Inn}(\mathfrak{g})\subset\operatorname{Aut}(\mathfrak{g})$, to be the subgroup generated by the pure inner automorphisms. So my question is equivalent to: is the condition generated by strictly necessary? The family of pure inner automorphisms isn't in general a subgroup by itself?
