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 Galois 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 $\operatorname{Aut}_X(Y')$. $\square$

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. Let $U \subseteq X$ be the dense open locus where $f$ is étale, let $V = f^{-1}(U)$, let $\bar y \to V$ be a geometric point with image $\bar x \to U$, and let $\phi \colon \pi_1(U,\bar x) \twoheadrightarrow G$ be the surjection corresponding to the $G$-cover $V \to U$. 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. The isomorphism $a$ takes $U$ to itself, and the pullback $V' \to U$ of $V \to U$ along $a$ is isomorphic to $V \to U$ (as étale $G$-covers of $U$);
4. For any choice of path $[\gamma] \in \pi_1(U,\bar x, a^*\bar x)$, the subgroups $\ker \phi$ and $\ker(\phi a_* \gamma_*)$ of $\pi_1(U,\bar x)$ are conjugate (see proof for precise statement).

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 $$\begin{array}{ccc}Y & \stackrel{b}\dashrightarrow & Y \\ \downarrow & & \downarrow \\ X & \underset a\to & X,\!\end{array}\tag{1}\label{1}$$ 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 (2) $\Leftrightarrow$ (3), we know that $b$ gives an isomorphism $V \to V$ lifting $a|_U \colon U \to U$. This is exactly the same thing as an isomorphism $V \to V'$ over $U$, where $V'$ is the pullback $$\begin{array}{ccc}V' & \to & V \\ \downarrow & & \downarrow \\ U & \stackrel a\to & U.\!\end{array}$$ Finally, for (3) $\Leftrightarrow$ (4), we note that the cover $V' \to U$ corresponds to the surjection $$\pi_1(U,a^*\bar x) \stackrel{a_*}\to \pi_1(U,\bar x) \stackrel \phi\twoheadrightarrow G.$$ Any choice of path $[\gamma] \in \pi_1(U,\bar x, a^*\bar x)$ gives an identification \begin{align*} \gamma_* \colon \pi_1(U,\bar x) &\stackrel\sim\longrightarrow \pi_1(U,a^* \bar x)\\ [\alpha] &\longmapsto [\gamma^{-1}] \cdot [\alpha] \cdot [\gamma], \end{align*} well-defined up to conjugation. Under this identification, the surjection $\pi_1(U,a^*\bar x) \twoheadrightarrow G$ above corresponds to the surjection $\pi_1(U,\bar x) \twoheadrightarrow G$ given by $\phi a_* \gamma_*$. The induced cover is isomorphic to the cover $V \to U$ given by $\phi$ if and only if the kernels are conjugate (see e.g. [Munkres, Thm. 79.4] in the topological setting), proving (3) $\Leftrightarrow$ (4).

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 $\operatorname{Aut}_X(Y')$. (See also this post for a general discussion of Galois covers of normal schemes.) $\square$

References.

[Munkres] J. R. Munkres, Topology (second edition). Pearson, 2018.

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 $\operatorname{Aut}_X(Y')$. $\square$

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 Galois 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 $\operatorname{Aut}_X(Y')$. $\square$