Let $G$ be a Lie group, $\operatorname{Aut}(G)$ the group of diffeomorphisms of $G$ that are also homomorphisms. Denote by $\operatorname{Inn}(G) \unlhd \operatorname{Aut}(G)$ the group of automorphisms which correspond to conjugation by some $h \in G$. Define $$\operatorname{Out}(G) := \operatorname{Aut}(G) / \operatorname{Inn}(G).$$ I would like to show that $$\operatorname{Out}(\operatorname{SL}_3(\mathbb{R}))\cong \mathbb{Z}/2 \mathbb{Z}.$$ I have already seen computations for different Lie groups, namely $\operatorname{SL}_2(\mathbb{C})$ and $\operatorname{SL}_2(\mathbb{R})$. The common idea seems to be to use the fact that if $\Phi: G \to G$ is an automorphism, then its differential at the identity $d_e \Phi: \mathfrak{g} \to \mathfrak{g}$ is an automorphism as well and then to use the structure on the corresponding Lie algebra.
I hardly know anything about the Lie algebra $\mathfrak{sl}_3(\mathbb{R})$. What would be a good approach to show the claim above? Can it be done in a similar way or is there an easier approach?
For the Lie algebra $\mathfrak{sl}_n(\mathbb{C})$ the inner automorphism group is isomorphic to $PGL_n(\mathbb{C})$, and the outer automorphism group is isomorphic to $\mathbb{Z}/2\mathbb{Z}$ for $n\ge 3$. This is proved in Jacobson's book on Lie algebras, chapter $IX$. Indeed, the outer automorphism different from the identity is simply given by $X\mapsto -X^T$. The result is proved for an arbitrary algebraically closed field of characteristic zero.