In this post I will work in the greatest generality I can think of. In particular, fundamental group means étale fundamental group, but for varieties over $\mathbf C$ the same argument carries through using the topological fundamental group instead.

> **Lemma.** *Let $X$ and $Y$ be separated normal integral schemes, let $f \colon Y \to X$ be a finite and finitely presented separable Galois cover with group $G$, and let $a \colon X \to X$ be an automorphism. Then the following are equivalent:*
> 1. *There exists an automorphism $b \colon Y \to Y$ lifting $a$;*
> 2. *There exists a dominant rational map $b \colon Y \to Y$ lifting $a$;*
> 3. *There exists an automorphism $\phi \colon G \to G$ and a commutative diagram*
$$\begin{array}{ccc}\pi_1(X) & \stackrel{a_*}\to & \pi_1(X) \\ \downarrow & & \downarrow \\ G & \underset{\phi}\to & G.\! \end{array}\tag{1}\label{1}$$
> *Moreover, the set of such lifts is a $G$-bitorsor via pre- and post-composition of deck transformations.*

*Proof.* For (1) $\Leftrightarrow$ (2), note that a dominant rational lift $b$ is automatically an automorphism. Indeed, given a commutative diagram
$$\begin{array}{ccc}Y & \stackrel{b}\dashrightarrow & Y \\ \downarrow & & \downarrow \\ X & \underset a\to & X,\!\end{array}\tag{2}\label{2}$$
multiplicativity of function field degrees shows that $b$ is birational. Since $Y$ is the integral closure of $X$ in $K(Y)$, we conclude that $b$ is an isomorphism since normalisation is a functor.

Thus, for (1) $\Leftrightarrow$ (3) we may restrict to the (nonempty open) étale locus to assume $f$ is étale. To give a diagram as in (\ref{2}) is equivalent to giving an isomorphism $Y \stackrel\sim\to Y'$ over $X$, where $Y' \to X$ is the base change of $Y \to X$ along $a$. By the theory of the fundamental group, this is equivalent to an isomorphism $\phi \colon G \stackrel \sim\to a^*G$ as groups under $\pi_1(X)$, where $a^*G$ denotes the map $\pi_1(X) \stackrel{a_*}\to \pi_1(X) \twoheadrightarrow G$. This is exactly the content of diagram (\ref{1}).

The final statement follows for example because $\operatorname{Isom}_X(Y,Y')$ is naturally a $G$-bitorsor, as $G$ agrees with both $\operatorname{Aut}_X(Y)$ and $\operatorname{Aut}_X(Y')$. $\square$