the title is quite self-explaining. Let's take a corona $\{z\in\mathbb{C},r<\mid z\mid<R\}$, for two strictly positive reals $r,R$. I am looking for a "classical" lie group that acts holomorphically transitively on the corona. And if not, a non classical will do the stuff;)
I thought of a subgroup of the isometries of the unit disk in the poincaré model, but I think there can not be any of this sort.
For a continuous action, the answer is yes: $\mathbb{C}-\{0\}=\mathrm{GL}(1,\mathbb{C})$ is a classical Lie group which acts transitively on itself by multiplication. Now use that $\{z\in\mathbb{C}:r<|z|<R\}$ is homeomorphic to $\mathbb{C}-\{0\}$.
For a holomorphic action, the answer is no: If $f$ is a biholomorphism of the annuli, then either $|f(z)|=|z|$ for all $z$ or $|f(z)|=Rr/|z|$ for all $z$ (assuming $0<r,R<\infty$) (see here). Hence, no holomorphic action can be transitive.