Let $\mathfrak{g}$ be a seimisimple Lie algebra over $\mathbb{C}$, $\mathfrak{h}$ a Cartan subalgebra, $\mathrm{Aut}^0(\mathfrak{g})$ the connected component of the identity in the automorphisms group of $\mathfrak{g}$. What I want to prove is there exists an isomorphism
$$\left \{\phi \in \mathrm{Aut}^0(\mathfrak{g}) \mid \phi(\mathfrak{h})=\mathfrak{h} \right \}/\left \{\phi \in \mathrm{Aut}^0(\mathfrak{g}) \mid \phi_{\mid \mathfrak{h}} = \mathrm{id}_{\mathfrak{h}} \right \} \cong \mathrm{Weyl \ group}.$$
Denote $N$ by $\left \{\phi \in \mathrm{Aut}^0(\mathfrak{g}) \mid \phi(\mathfrak{h})=\mathfrak{h} \right \}$. It is obvious that every element in $\mathrm{Aut}(\mathfrak{g})$ preserves the Killing form of $\mathfrak{g}$, in particular, every element in $N$ as well and for every element $\phi$ in $N$, we have $\phi(\alpha^*) = (\alpha \circ \phi^{-1})^{*}$ where $\alpha^*$ is the dual of an eigenform, i.e. $\alpha(H) = B(H,\alpha^*)$ ($B$: the Killing form). On the real vector space $E = \sum \mathbb{R}\alpha^*$ we have
$$B(\phi X,Y) = B(X,\phi^{-1}Y),$$
therefore with respect to the Killing form as a quadratic form, elements in $N$ take value in $O(E)$ – the orthogonal group, i.e. there exists a canonical map
$$\begin{align*} \psi: N & \to O(E) \\ \phi & \mapsto \left \{\phi: \alpha^* \mapsto \phi(\alpha^*) \right \} \end{align*},$$
and this map is factored through $N/\left \{\phi \in \mathrm{Aut}^0(\mathfrak{g}) \mid \phi_{\mid \mathfrak{h}}=\mathrm{id}_{\mathfrak{h}} \right\}$ since $\left \{\phi \in \mathrm{Aut}^0(\mathfrak{g}) \mid \phi_{\mid \mathfrak{h}}=\mathrm{id}_{\mathfrak{h}} \right\}$ is precisely the kernal of $\psi$; in particular, the induced map is injective. If we can prove that this map takes value in the Weyl group then we deduce an isomorphism. Indeed, this map is surjective for each reflection $S_{\alpha}$ because in $\mathfrak{g}$, the subalgebra $g_{\alpha} \oplus g_{-\alpha} \oplus [g_{\alpha},g_{-\alpha}]$ is isomorphic to $\mathfrak{sl}_2$ and we can choose element $X,H,Y$ in this subalgebra such that they correspond to standard basis of $\mathfrak{sl}_2$, a direct computation shows that $\psi(e^{\mathrm{ad}(X)}e^{\mathrm{ad}(-Y)}e^{\mathrm{ad}(X)}) = S_{\alpha}$. What is left here is the fact that the induced map of $\psi$ takes values in the Weyl group.
From the semisimplicity, $\mathrm{Aut}^0(\mathfrak{g})$ is the inner automorphism group $\left \langle e^{\mathrm{ad}(x)}\mid x \in \mathfrak{g} \right \rangle$ but I do not know how to proceed.
Best Answer
This is part of proposition 4 in Bourbaki's books on Lie Groups and Algebras, ch. VIII ยง5 no. 2. Since Bourbaki's setting is slightly more general (they are looking at a semisimple $\mathfrak g$ with a splitting Cartan subalgebra $\mathfrak h$ over a general characteristic $0$ field $k$, where the definition of $Aut^0$ is those automorphisms which become elementary (i.e. products of automorphisms of the form $e^{ad(x)}$) after scalar extension to an algebraic closure), and their overall statement is also more general, it is a bit hard to track down what exactly we would need for the proof of the inclusion $\psi(N) \subseteq W$ you are after. All the more so since, as usual, Bourbaki's proof can lead us down multiple rabbitholes of references and lemmata before (in particular lemma 2 before the proposition seems to be crucial here). But I think the approach is:
As you say in a comment, via twisting $\phi \in N$ with an element which lands in the Weyl group, it suffices to show that if $\psi(\phi)$ stabilizes (as a set) a base $B$ of our root system $R$, then it must be the identity on $R$ (i.e. fix it pointwise).
So assume we have such a $\phi$. We can now twist it further, without changing its image under $\psi$, via certain elementary automorphisms (which act as identity on $\mathfrak h$ but scale the elements of certain root spaces) to ensure that on the entire space $\sum_{\alpha \in R} \mathfrak g_\alpha$ (the sum of all root spaces, a vector space complement of $\mathfrak h$ in $\mathfrak g$), $\phi$ does not have the eigenvalue $1$. (The exact construction of such a twisting takes up most of Bourbaki's proof of the proposition and looks very technical at first, but the idea is kind of straightforward.)
Now the above mentioned lemma 2 says that this last statement implies that $\phi$ is the identity on $\mathfrak h$. To see this, they use that on the one hand, the nilspace of $\phi -id$ in $\mathfrak g$ has dimension $\ge \dim\mathfrak h$ (that comes from a general fact about CSA's as minimal Engel subalgebras proven in an earlier volume), so by that assumption of no eigenvalue $1$ on that complement, this nilspace is $\mathfrak h$, i.e. $\phi-id$ is nilpotent on $\mathfrak h$; on the other hand, because the restriction of $\phi$ to $\mathfrak h$ has finite order (it stabilizes the finite set of coroots, which span $\mathfrak h$), by Maschke's Theorem it is semisimple (in our case $k=\mathbb C$, diagonalizable, and there might be an easier way to see this); putting this together, when restricted to $\mathfrak h$, the endomorphism $\phi - id$ is semisimple and nilpotent, hence identically zero.