An automorphism of the form $\exp(\operatorname{ad}x)$, $\operatorname{ad}x$ nilpotent, is called inner; more generally, the subgroup of $\operatorname{Aut}L$ generated by these is denoted $\operatorname{Int}L$ and its elements called inner automorphisms. It is a normal subgroup: If $\phi\in\operatorname{Aut}L, x\in L$, then $\phi(\operatorname{ad}x)\phi^{-1}=\operatorname{ad}\phi(x)$, whence $\phi\exp(\operatorname{ad}x)\phi^{-1}=\exp(\operatorname{ad}\phi(x))$.
I don't quite understand why $\phi(\operatorname{ad}x)\phi^{-1}=\operatorname{ad}\phi(x)$. Suppose the left-hand side is applied on $y\in L$, then we get $\phi(\operatorname{ad}x)\phi^{-1}y = \phi[x, \phi^{-1}y]$, while if the right-hand side is applied on $y\in L$, then we get $[\phi(x), y]$.
We don't have $\phi([x, \phi^{-1}(y)])=[\phi(x), y]$, do we?
Sure. An automorphism of a Lie algebra respects all of its structure, including the bracket. So for $\phi\in\operatorname{Aut}L$ and $x,y\in L$ we have $\phi([x,y])=[\phi(x),\phi(y)]$.