A question on Aut$(N)$ and Aut$(N/G)$ Let $N$ be a complex manifold and $G$ is a finite group freely acting on $N$. Define another complex manifold $M$ as $M=N/G$. I would like to study Aut$(M)$, the (holomorphic) automorphism group of $M$. 
Is it true that Aut$(M)=N_G($Aut($N))$? Here $N_G($Aut($N))$ denotes the normalizaer of $G$ inside Aut($N)$. 
 A: Claim: If $N$ is a holomorphic manifold, $G \subseteq \text{Aut}(N)$ is a finite group acting freely on $N$, and $M = N/G$ is the quotient manifold, then $\text{Aut}(M) \simeq N_G\bigl(\text{Aut}(N)\bigr)/G$, the quotient of the normalizer of $G$ in $\text{Aut}(N)$ by $G$.
Proof: There is a homomorphism $\phi:N_G\bigl(\text{Aut}(N)\bigr)/G \to \text{Aut}(M)$ since (i) an automorphism $f$ of $N$ induces an automorphism on the quotient if and only if $f$ normalizes $G$, and (ii) $\ker(\phi) = G$, i.e., $f$ induces the identity on $M = N/G$ if and only if $Gf = G$, if and only if $f \in G$.
To prove $\phi$ is surjective, let $\pi:N \to M = N/G$ denote the quotient map, and let $\bar{f}$ be an automorphism of $M$. Since $\pi$ is an unramified finite-sheeted covering, the map $\bar{f} \circ \pi:N \to M$ lifts to a holomorphic bijection $f:N \to N$. Since the inverse of a holomorphic bijection is itself holomorphic, $f \in \text{Aut}(N)$, and $\phi(f) = \bar{f}$ by construction.
[This sketch is marked "community wiki" in case it's perceived to be lacking in detail or scope; e.g., I haven't addressed studiosus' comment about other structures or made any attempt to weaken hypotheses. --Andy]
