Every automorphism preserves a Cartan subalgebra is induced by an element in the Weyl group 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.
 A: This is part of proposition 4 in Bourbaki's books on Lie Groups and Algebras, ch. VIII §5 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.
A: Since the Bourbaki style is terrible to track down all the detals, I post this one to help others those who want to read a full proof of this problem; of course, I adapt modern notations.
Fix a base $\Delta$ for the root system $\Sigma$. Denote
$$\Sigma^* = \left \{\alpha^* \mid \alpha \ \text{is a root} \ \right \}$$
$$\mathrm{Aut}(\Sigma^*,\Delta) = \left \{ \ \text{elements in} \ \mathrm{Aut}(\Sigma^*) \ \text{stabilize} \ \Delta \right \}.$$
We are now at the position to prove that if $\phi \in N$ s.t. $\psi(\phi)$ induces an element in $\mathrm{Aut}(\Sigma^*,\Delta)$ then $\delta=\psi(\phi)$ is induced by an element in $Z = \left \{\phi \in \mathrm{Aut}_0(\mathfrak{g}) \mid \phi_{\mid \mathfrak{h}} = \mathrm{id}_{\mid \mathfrak{h}} \right \}$. The subgroup generated by $\delta$ has a finite number of orbits on $\Sigma^*$, let $U$ be such an orbit of cardinal $r$, denote
$$g_U = \bigoplus_{\alpha^* \in U} g_{\alpha}.$$
Let $\alpha_1^*\in U$ and I define $\alpha_i^* = \delta^{i-1}(\alpha_1^*) = (\alpha_1 \circ \delta^{1-i})^* \ \forall \ i = \overline{1,r}$. Thus $U = \left \{\alpha_1^*,...,\alpha_r^* \right \}$. Let $X_1$ b a non-zero element in $g_{\alpha_1}$. We are going to prove that there exists a non-zero scalar $c_U$ such that $\delta^r(X_1)=c_U X_1$. To do this, recall that $U$ is an orbit, therefore $\delta^r(\alpha_1^*) = \alpha_1^*$, equivalently, $(\alpha_1 \circ \delta^{-r})^* = \alpha_1^*$. We claim that $\delta^r(X_1) \in g_{\alpha_1}$. Indeed, for all $H \in \mathfrak{h}$,
$$\begin{align*}
[H,\delta^r(X_1)] & = \delta^r[\delta^{-r}(H),X_1] \\ & = \delta^r\left((\alpha_1\circ \delta^{-r})(H)X_1 \right) \\& = (\alpha_1\circ\delta^{-r})(H) \delta^r(X_1) \\ & = B(H,(\alpha_1 \circ \delta^{-r})^*) \delta^r(X_1) \\ & = B(H,\alpha_1^*)\delta^r(X_1) \\ & = \alpha_1(H)\delta^r(X_1). \end{align*}$$
But by the semisimplicity of $\mathfrak{g}$, $\mathrm{dim}(g_{\alpha})=1$, hence $\delta^r(X_1)$ and $X_1$ are proportional, prove our claim. By the definition of other $X_i$, we deduce that
$$\delta^r_{\mid g_U} = c_U.\mathrm{id}_{g_U}.$$
We shall twist $\delta$ by an element of $Z$ so that the resulting automorphism still comes from an element in $Z$. For each functional,
$$\Theta: \bigoplus_{\alpha \in \Sigma} \mathbb{Z}\alpha^* \to \mathbb{C}^*$$
We define an element $Z$, denoted by $f(\Theta)$, by the following rule
$$\begin{cases} f(\Theta)_{\mid g_{\alpha}} = \Theta(\alpha^*)\mathrm{id}_{\alpha} \\ f(\Theta)_{\mid \mathfrak{h}} = \mathrm{id}_{\mathfrak{h}} \end{cases}$$
It is clear that $f(\Theta) \in Z$. Moreover,
$$(\delta \circ f(\Theta))(X_1) = \delta(\Theta(\alpha_1^*)X_1) = \Theta(\alpha_1^*)\delta(X_1) = \Theta(\alpha_1^*)X_2.$$
Iterating this process by successively applying $\delta \circ f(\Theta)$, we deduce that
$$(\delta \circ f(\Theta))^r(X_1) = c_U\prod_{i=1}^r \Theta(\alpha_i^*)X_1 = c_U \Theta \left(\sum_{i=1}^r \alpha_i^* \right)X_1.$$
Again, this implies that
$$(\delta \circ f(\Theta))^r_{\mid g_U} = c_U \Theta \left(\sum_{i=1}^r \alpha_i^* \right) \mathrm{id}_{g_U}.$$
Let write our fixed base $\Delta$ as $\Delta = \left \{\beta_1^*,...,\beta_n^* \right \}$. Since this is a base, there exists $a_1^U,...,a_1^U \in \mathbb{Z}$, have same sign, not all zero such that $\alpha_1^*= \sum_{i=1}^n a_i^U \beta_i^*$, successively apply $\delta$ and recall that $\delta \in \mathrm{Aut}(\Sigma^*,\Delta)$ we deduce the existence of $m_1^U,...,m_n^U \in \mathbb{Z}$ of same sign, not all zero such that
$$\sum_{i=1}^r \alpha_i^* = \sum_{i=1}^n m_i^U \beta_i^*.$$
Thus, define $c'_U =  c_U\Theta \left(\sum_{i=1}^r \alpha_i^* \right) = c_U \prod_{i=1}^n \Theta(\beta_i^*)^{m_i^U}$, we can choose $\Theta$ such that for all orbits $U$, $c_U' \neq 1$. This is possible, because it is equivalent to choose $\Theta(\beta_i^*) = t_i$ are elements in $\mathbb{C}^*$ such that all polynomial $c_U\prod_{i=1}^n t_i^{m_i^U} - c'_U$ vanish. So far, what we've done is to show that it is possible to choose $\Theta$ such that $\delta \circ f(\Theta)$ does not have $1$ as an eigenvalue on the subspace $\oplus_{\alpha \in \Sigma} g_{\alpha}$.
Now since $\mathrm{Aut}^0(\mathfrak{g})$ is a connected Lie group, every element can be written as product of some $e^{y}$ with $y \in \mathrm{Lie}(\mathrm{Aut}^0(\mathfrak{g})) \subset \mathrm{Der}(\mathfrak{g}) \overset{\text{semisimplicity}}{=} \mathrm{ad}(\mathfrak{g})$; thus, each element in $\mathrm{Aut}^0(\mathfrak{g})$ is a product of elements of form $e^{\mathrm{ad}(x)}$ with $x \in \mathfrak{g}$. We are at the position to prove that the generalized $0$-space of $\delta \circ f(\Theta) - \mathrm{id}$ has dimension at least the dimension of $\mathfrak{h}$, which equals $\mathrm{rank}(\mathfrak{g})$. To illustrate how to do this, we consider the case $\delta \circ f(\Theta) = e^{\mathrm{ad}(x)}$ only. Since we are working over $\mathrm{C}$, $\mathrm{ad}(x)$ is similar to a Jordan matrix $J$,  read $\mathrm{ad}(x) = AJA^{-1}$ for some $A \in \mathrm{GL}(\mathfrak{g})$. Thus, $\mathrm{dim} \mathfrak{g}^0(\mathrm{ad}(x))$ is precisely the number of $0$ on the diagonal. Moreover, $e^{\mathrm{ad}(x)} - \mathrm{id} = A(e^J - 1)A^{-1}$ and $e^{\lambda}=1 \Leftrightarrow \lambda=0$, we conclude that the number of $0$ on the diagonal of $e^J - \mathrm{id}$ equals of $\mathrm{ad}(x)$. Finally
$$\mathrm{dim} \bigcup_{k \geq 1} (\delta \circ f(\Theta) - \mathrm{id})^k = \mathrm{dim} \mathfrak{g}^0(\mathrm{ad}(x)) \geq \mathrm{rank}(\mathfrak{g}) = \mathrm{dim}\mathfrak{h}.$$
By the assumption of no eigenvalue $1$ on the complement $\oplus_{\alpha \in \Sigma}g_{\alpha}$, we see that $\delta \circ f(\Theta) - \mathrm{id}$ is nilpotent on $\mathfrak{h}$. On the other hand, $\delta \circ f(\Theta)$ permutes a base so it has finite order (it is an element of a finite symmetric group). Consequently, its characteristic polynomial (and hence minial polynomial) divides $x^k-1$ for some $k$, the latter polynomial has all roots as simple roots so the Jordan blocks of $\delta \circ f(\Theta)$ are all of size $(1 \times 1)$; i.e. $\delta \circ f(\Theta)$ is diagonalizable. Combine this with the previous assertion of $\delta \circ f(\Theta)$ being nilpotent, we deduce that $\delta \circ f(\Theta)_{\mid \mathfrak{h}} = \mathrm{id}_{\mathfrak{h}}$, as desired.
