I'm studying Lie Algebras and Lie Groups from the book "Lie algebras and Lie groups" by Bourabki Nicolas. There he claims that if $H$ is a connected closed subgroup of a Lie group $G$, then the normalizer of $L(H)$ (the Lie algebra of H) is exactly $L(N_G(H))$ - that is the Lie algebra of the normalizer of $H$ in $G$. This proof I understand, but I have a different question - if $X$ is an element of the normalizer of $L(H)$, is it true that: $$DAd_{\exp(X)}(L(H))=L(H)$$ with $DAd_{\exp(X)}$ being the differential of $Ad_{\exp(X)}:G\to G$?
I know that $[X,Y]=(\frac{d}{dt}|_{t=0}Ad_{\exp(tX)})(Y)$, but can't see if it implies the former.
Any help would be appreciated.
Looking at my comment again I think it could be made a little clearer so I'll turn it into an answer.
The first thing to note is that you have called two different maps $\mathrm{Ad}$ here so let's make that clearer.
Let $g \in G$ and $X\in L(G)$. We can define an automorphism $C_g:G \to G$ by conjugation $C_g(h) = ghg^{-1}$. Note we always have $C_g(e)=e$ so this differentiates to an action on $L(G) = T_eG$. This is differentiation in the sense of a pushforward of a map between manifolds. We call this map $\mathrm{Ad}_g$ and it turns out to be an automorphism of $L(G)$. The map $\mathrm{Ad}:G \to \operatorname{GL}(L(G));g\mapsto \mathrm{Ad}_g$ is precisely what we call the adjoint representation of $G$.
We can differentiate this map $\mathrm{Ad}$ (note no choice of $g$ now) at $e$ to define a map $\mathrm{ad}:L(G) \to \mathrm{End}(L(G))$ from which it follows that $\left.\frac{d}{dt}\right|_{t=0}\mathrm{Ad}_{\exp(tX)} = \mathrm{ad}(X)$. We can also write $\mathrm{ad}(X)(Y) = [X,Y]$ as this coincides exactly with Lie bracket induced from the definition via left-invariant vector fields.
So in this notation, what you call $D\mathrm{Ad}_g$ is actually just $\mathrm{Ad}_g$ and we should call it $DC_g$ instead. It is easy to mix these up as in matrix groups they are both conjugation (and so are often denoted the same way) but one is acting on Lie group elements and one is acting on Lie algebra elements.
In this framing, you have already answered the question in the first part since you already know that the normaliser of $L(H)$ in the Lie group and Lie algebra correspond so that $\exp$ must map between them and so $\mathrm{Ad}_{\exp(X)}(L(H)) \subset L(H)$ for any $X$ normalising $L(H)$.
You could also see this by the handy relation $\mathrm{Ad}_{\exp(X)} = \exp(\mathrm{ad}(X))$ where $$ \exp(\mathrm{ad}(X))(Y) = Y + [X,Y] + \frac{1}{2!}[X,[X,Y]] + \frac{1}{3!}[X,[X,[X,Y]]] + \dotsm$$