How do we construct a precise map of inner + outer automorphism of special orthogonal group $SO(n;\mathbb{R})$?
$d=2$; We can look at $SO(2;\mathbb{R})=U(1)$ which is abelian, and we know the inner $$ Inn(SO(2;\mathbb{R}))=SO(2;\mathbb{R})/Z(SO(2;\mathbb{R}))=1 $$ $$ Out(SO(2;\mathbb{R}))=\mathbb{Z}/2 $$ The total $Aut(SO(2;\mathbb{R}))=Inn(SO(2;\mathbb{R})) \rtimes Out(SO(2;\mathbb{R}))=\mathbb{Z}/2$ We have no $Inn(SO(2;\mathbb{R}))$ except the identity map. I believe that we can get the $Out(SO(2;\mathbb{R}))=\mathbb{Z}/2$ by flipping $t \to -t$ in $$ U(1)=\{\exp(i t) | t \in [0, 2 \pi)\} \to \{\exp(-i t) | t \in [0, 2 \pi)\}. $$ I wish to see explicit answer like the above for my following questions ---
other $n$ but $n\neq 2,8$,
Question 1: How do we construct the inner automorphism map explicitly (if my result is correct?) $$ Inn(SO(n;\mathbb{R}))=SO(n;\mathbb{R})/Z(SO(n;\mathbb{R})) =\left\{ \begin{array}{ll} SO(n;\mathbb{R})/\mathbb{Z}/2, &n \in even\\ SO(n;\mathbb{R}) , &d \in odd \end{array} \right. $$
Question 2: How do we construct the outer automorphism map explicitly $$ Out(SO(n;\mathbb{R}))=\mathbb{Z}/2 $$ Given the parametrization of $SO(n;\mathbb{R})$ how to map to itself via the Out map?
Question 3: How do we construct the total automorphism map explicitly $$ Aut(SO(n;\mathbb{R}))=Inn(SO(n;\mathbb{R})) \rtimes Out(SO(n;\mathbb{R}))? $$ Given the parametrization of $SO(n;\mathbb{R})$ how to map to itself via the Aut map?
p.s. Possible useful link but with not explicit (not enough) constructions in https://mathoverflow.net/questions/235758/automorphism-group-of-real-orthogonal-lie-groups