Automorphisms of the Lie algebras $\mathfrak{sl}(2,R)$ and $\mathfrak{su}(2)$ I would like to know about the literature concerning the group of outer automorphisms of the Lie algebra $\mathfrak{sl}(2,R)$. This question is addressed in different places in a contradictory way. In certain works, e.g. 
M.A. Farinati and A.P Jancsa,
Three dimensional real Lie bialgebras,
Revista de la union matematica argentina Vol. 56, No. 1, 2015, Pages 27–62, 2015,
it is implicitly claimed that the aforesaid group is trivial and all automorphisms of $\mathfrak{sl}(2,R)$ are inner. By googling the question, I also found several sources claiming that the group of inner automorphisms of $\mathfrak{sl}(2,R)$ is PSL(2,R) (see https://groupprops.subwiki.org/wiki/Special_linear_group:SL(2,R)), and the outer automorphisms are given by PGL(2,R) (see Outer automorphisms of simple Lie Algebras) which are obviously different.
Similarly, I would like to know about the structure of the group of outer automorphisms of the Lie algebra $\mathfrak{su}(2)$
Thank you in advance for your comments.
 A: It is clear that $PGL(2,\mathbb{R})$ acts as automorphisms. It is easy to check that the reflections act in a manner unlike any positive determinant matrices. Hence the automorphism group is larger than $PSL(2,\mathbb{R})$. Since we know the answer over $\mathbb{C}$ (as in Fulton and Harris, Representation Theory, p. 498), by complexification, we know that all automorphisms arise from conjugation by some matrices. We can easily check that complex matrices give us real automorphisms only when they are real up to a constant scaling. So we see that the automorphism group of $\mathfrak{sl}(2,\mathbb{R})$ is $PGL(2,\mathbb{R})$.
For $\mathfrak{su}(2)$, any automorphism complexifies to an automorphism of $\mathfrak{sl}(2,\mathbb{C})$, so arises by conjugation from a complex matrix $g$, again from Fulton and Harris. To get conjugation by $g$ to preserve the real subspace $\mathfrak{su}(2)$, we need $gAg^{-1}$ to be special unitary for any special unitary $A$. Plug this in and check that this forces $g^*g$ to commute with all such $A$, so by Schur's lemma, $g^* g=\lambda I$ for some complex number $\lambda$. Take determinant to find that $\lambda=\pm 1$. Replace $g$ by $ig$ if needed to get $\lambda=1$, so $g \in SU(2)$. Check that $g$ acts trivially if and only if $g=-I$ to see that the automorphism group of $\mathfrak{su}(2)$ is $SO(3)=SU(2)/\pm 1$.
