Deck transformation group in algebraic geometry

Solution 1:

I am sorry that I made some mistakes. This is not true, even for smooth curves. For example, let $X$ be a general Riemann surface with genus at least $3$. Then it does not have non-trivial automorphism, but there exists meromorphic functions, hence a branched cover $X\to Y$. In particular, ${\rm Aut}(X/Y)$ is trivial. However, the monodromy group is clearly not trivial.

In general, let ${\rm Gal}(K(Y)/K(X))$ be the automorphism group of the Galois closure, then

$${\rm Aut_{rational}}(X/Y)\cong{\rm Aut}(K(X)/K(Y))$$ and $$G \cong {\rm Gal}(K(X)/K(Y)).$$

For smooth curves, ${\rm Aut}(X/Y)={\rm Aut_{rational}}(X/Y)$. Then a sufficient and necessary condition to ensure $G\cong {\rm Aut(X/Y)}$ is that the covering $X\to Y$ is Galois.