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 $SO(3)=SU(2)/\pm 1$.