Assume that $M$ is a complex manifold.


Let $G$ be the group of all (real) smooth diffeomorphisms $\phi$ of $M$ such that $\phi^\* (X)$ is a holomorphic vector field for all holomorphic vector fields $X$ on $M$. Is $G$ a finite dimensional Lie group?(With respect to a natural smooth structure on $G$).



Real diffeomeorphism preserving the space of Holomorphic vector fields
Take complex vector space $V$, say of complex dimension $n$. Take a complex linear map $A \colon V \to V$ whose eigenvalues $\lambda$ all satisfy $|\lambda|>1$. The group generated by $A$ acts on $V-0$ freely and properly, with quotient a compact complex manifold $M=(V-0)/(z \sim Az)$, called a *primary Hopf manifold*. The group of linear transformations commuting with $A$ acts on $V$, transitively on a dense open set. These linear transformations descend to biholomorphisms of $M$. Any holomorphic vector field on $M$ lifts to a holomorphic vector field $v \colon V-0 \to V$ commuting with $A$. By Hartogs extension theorem, $v$ extends to a holomorphic map $v \colon V \to V$. Expand in a Taylor series, and apply $A$-invariance, to prove that $v$ is linear. Hence the biholomorphism group of $M$ is the group of linear transformations commuting with $A$. This group preserves the generalized eigenspaces of $A$, so these project to submanifolds of $M$ which are invariant under all holomorphic vector fields. Your group $G$ preserves the holomorphic vector fields, and therefore preserves their orbits, the images of the generalized eigenspaces. So $G$ is not compact, since it contains the invertible linear transformations commuting with $A$. But $G$ acts transitively only on a dense open set, and not on all of $M$.


