In this delightful question, the poster mentioned that the isometry group of a compact Lie group $G$, equipped with the metric from the Killing form, is $G\times G/Z(G)$, where $Z(G)$ is the center of $G$. I'm specifically interested in the Riemannian case, so I will add the extra condition that $G$ be semisimple.
Is there a concise way to show that $G\times G/Z(G)$ is the isometry group, other than using the algorithm from the answer?
I can clearly see that it is a subgroup, since the metric coming from the Killing form is bi-invariant, and that quotienting out the center from one of the two copies of $G$ is necessary, because otherwise the action wouldn't be faithful. However, I cannot see how all isometries are of that form. In fact, I'm almost positive that this would exclude something of finite order, like $g\mapsto g^{-1}$.