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It seems to me that the answer is no even if the action of $G$ is free.

In fact, assume that $G$ acts freely and set $Y:= X/G$. Let $f \colon Y \to Y$ be an automorphism and $f_* \colon \pi_1(Y) \to \pi_1(Y)$ the induced automorphism of the fundamental group. The Galois cover $p \colon X \to Y$ is induced by a normal subgroup of $\pi_1(Y)$, that we call $H$, such that $\pi_1(Y)/H = G$. It is immediate to check that

if $f \colon Y \longrightarrow Y$ lifts to $g \colon X \longrightarrow X$, then necessarily $f_*H=H$.

Now let me give an example of a free $G$-action and an automorphism that does not lift.

Let $Y = \mathbf{C}/ \Lambda$, where $\Lambda = \mathbf{Z} \oplus i \mathbf{Z}$. Then it is well-known that the elliptic curve $Y$ has an automorphism $f \colon Y \to Y$ of order $4$, induced by $z \to iz$.

Now set $X = \mathbf{C}/\Gamma$, where $\Gamma = \mathbf{Z} \oplus 2i \mathbf{Z}$. Since $\Gamma$ has index $2$ in $\Lambda$, we have an étale double cover (degree $2$ isogeny) $p \colon X \to Y$.

However, the automorphism $f \colon Y \to Y$ does not lift to $X$. The point is that the lattice $\Gamma$ is not invariant under $f_*$, in fact $f_*(\mathbf{Z} \oplus 2i\mathbf{Z}) = 2\mathbf{Z} \oplus i\mathbf{Z}.$

Remark. The same construction works for covers of any degree taking $\Gamma_n:= \mathbf{Z} \oplus ni\mathbf{Z}$. This lattice has index $n$ in $\Lambda$, hence there is a degree $n$ cyclic cover $p_n \colon X_n \to Y$, where $X_n = \mathbf{C}/ \Gamma_n$. The same argument used above shows that the automorphism $f \colon Y \to Y$ does not lift to $X_n$.

It seems to me that the answer is no even if the action of $G$ is free.

In fact, assume that $G$ acts freely and set $Y:= X/G$. Let $f \colon Y \to Y$ be an automorphism and $f_* \colon \pi_1(Y) \to \pi_1(Y)$ the induced automorphism of the fundamental group. The Galois cover $p \colon X \to Y$ is induced by a normal subgroup of $\pi_1(Y)$, that we call $H$, such that $\pi_1(Y)/H = G$. It is immediate to check that

if $f \colon Y \longrightarrow Y$ lifts to $g \colon X \longrightarrow X$, then necessarily $f_*H=H$.

Now let me give an example of a free $G$-action and an automorphism that does not lift.

Let $Y = \mathbf{C}/ \Lambda$, where $\Lambda = \mathbf{Z} \oplus i \mathbf{Z}$. Then it is well-known that the elliptic curve $Y$ has an automorphism $f \colon Y \to Y$ of order $4$, induced by $z \to iz$.

Now set $X = \mathbf{C}/\Gamma$, where $\Gamma = \mathbf{Z} \oplus 2i \mathbf{Z}$. Since $\Gamma$ has index $2$ in $\Lambda$, we have an étale double cover (degree $2$ isogeny) $p \colon X \to Y$.

However, the automorphism $f \colon Y \to Y$ does not lift to $X$. The point is that the lattice $\Gamma$ is not invariant under $f_*$, in fact $f_*(\mathbf{Z} \oplus 2i\mathbf{Z}) = 2\mathbf{Z} \oplus i\mathbf{Z}.$

Remark. A similar construction works for covers of any degree taking $\Gamma_n:= \mathbf{Z} \oplus ni\mathbf{Z}$. This lattice has index $n$ in $\Lambda$, hence there is a degree $n$ cyclic cover $p_n \colon X_n \to Y$, where $X_n = \mathbf{C}/ \Gamma_n$. The same argument used above shows that the automorphism $f \colon Y \to Y$ does not lift to $X_n$.

It seems to me that the answer is no even if the action of $G$ is free.

In fact, assume that $G$ acts freely and set $Y:= X/G$. Let $f \colon Y \to Y$ be an automorphism and $f_* \colon \pi_1(Y) \to \pi_1(Y)$ the induced automorphism of the fundamental group. The Galois cover $p \colon X \to Y$ is induced by a normal subgroup of $\pi_1(Y)$, that we call $H$, such that $\pi_1(Y)/H = G$. It is immediate to check that

if $f \colon Y \longrightarrow Y$ lifts to $g \colon X \longrightarrow X$, then necessarily $f_*H=H$.

Now let me give an example of a free $G$-action and an automorphism that does not lift.

Let $Y = \mathbf{C}/ \Lambda$, where $\Lambda = \mathbf{Z} \oplus i \mathbf{Z}$. Then it is well-known that the elliptic curve $Y$ has an automorphism $f \colon Y \to Y$ of order $4$, induced by $z \to iz$.

Now set $X = \mathbf{C}/\Gamma$, where $\Gamma = \mathbf{Z} \oplus 2i \mathbf{Z}$. Since $\Gamma$ has index $2$ in $\Lambda$, we have an étale double cover (degree $2$ isogeny) $p \colon X \to Y$.

However, the automorphism $f \colon Y \to Y$ does not lift to $X$. The point is that the lattice $\Gamma$ is not invariant under $f_*$, in fact $f_*(\mathbf{Z} \oplus 2i\mathbf{Z}) = 2\mathbf{Z} \oplus i\mathbf{Z}$

It seems to me that the answer is no even if the action of $G$ is free.

In fact, assume that $G$ acts freely and set $Y:= X/G$. Let $f \colon Y \to Y$ be an automorphism and $f_* \colon \pi_1(Y) \to \pi_1(Y)$ the induced automorphism of the fundamental group. The Galois cover $p \colon X \to Y$ is induced by a normal subgroup of $\pi_1(Y)$, that we call $H$, such that $\pi_1(Y)/H = G$. It is immediate to check that

if $f \colon Y \longrightarrow Y$ lifts to $g \colon X \longrightarrow X$, then necessarily $f_*H=H$.

Now let me give an example of a free $G$-action and an automorphism that does not lift.

Let $Y = \mathbf{C}/ \Lambda$, where $\Lambda = \mathbf{Z} \oplus i \mathbf{Z}$. Then it is well-known that the elliptic curve $Y$ has an automorphism $f \colon Y \to Y$ of order $4$, induced by $z \to iz$.

Now set $X = \mathbf{C}/\Gamma$, where $\Gamma = \mathbf{Z} \oplus 2i \mathbf{Z}$. Since $\Gamma$ has index $2$ in $\Lambda$, we have an étale double cover (degree $2$ isogeny) $p \colon X \to Y$.

However, the automorphism $f \colon Y \to Y$ does not lift to $X$. The point is that the lattice $\Gamma$ is not invariant under $f_*$, in fact $f_*(\mathbf{Z} \oplus 2i\mathbf{Z}) = 2\mathbf{Z} \oplus i\mathbf{Z}.$

Remark. The same construction works for covers of any degree taking $\Gamma_n:= \mathbf{Z} \oplus ni\mathbf{Z}$. This lattice has index $n$ in $\Lambda$, hence there is a degree $n$ cyclic cover $p_n \colon X_n \to Y$, where $X_n = \mathbf{C}/ \Gamma_n$. The same argument used above shows that the automorphism $f \colon Y \to Y$ does not lift to $X_n$.

It seems to me that the answer is no even if the action of $G$ is free.

In fact, assume that $G$ acts freely and set $Y:= X/G$. Let $f \colon Y \to Y$ be an automorphism and $f_* \colon \pi_1(Y) \to \pi_1(Y)$ the induced automorphism of the fundamental group. The Galois cover $p \colon X \to Y$ is induced by a normal subgroup of $\pi_1(Y)$, that we call $H$, such that $\pi_1(Y)/H = G$. It is immediate to check that

if $f \colon Y \longrightarrow Y$ lifts to $g \colon X \longrightarrow X$, then necessarily $f_*(H)=H$.

Now let me give an example of a free $G$-action and an automorphism that does not lift.

Let $Y = \mathbf{C}/ \Lambda$, where $\Lambda = \mathbf{Z} \oplus i \mathbf{Z}$. Then it is well-known that the elliptic curve $Y$ has an automorphism $f \colon Y \to Y$ of order $4$, induced by $z \to iz$.

Now set $X = \mathbf{C}/\Gamma$, where $\Gamma = \mathbf{Z} \oplus 2i \mathbf{Z}$. Since $\Gamma$ has index $2$ in $\Lambda$, we have an étale double cover (degree $2$ isogeny) $p \colon X \to Y$.

However, the automorphism $f \colon Y \to Y$ does not lift to $X$. The point is that the lattice $\Gamma$ is not invariant under $f_*$, in fact $f_*(\mathbf{Z} \oplus 2i\mathbf{Z}) = 2\mathbf{Z} \oplus i\mathbf{Z}$

It seems to me that the answer is no even if the action of $G$ is free.

In fact, assume that $G$ acts freely and set $Y:= X/G$. Let $f \colon Y \to Y$ be an automorphism and $f_* \colon \pi_1(Y) \to \pi_1(Y)$ the induced automorphism of the fundamental group. The Galois cover $p \colon X \to Y$ is induced by a normal subgroup of $\pi_1(Y)$, that we call $H$, such that $\pi_1(Y)/H = G$. It is immediate to check that

if $f \colon Y \longrightarrow Y$ lifts to $g \colon X \longrightarrow X$, then necessarily $f_*H=H$.

Now let me give an example of a free $G$-action and an automorphism that does not lift.

Let $Y = \mathbf{C}/ \Lambda$, where $\Lambda = \mathbf{Z} \oplus i \mathbf{Z}$. Then it is well-known that the elliptic curve $Y$ has an automorphism $f \colon Y \to Y$ of order $4$, induced by $z \to iz$.

Now set $X = \mathbf{C}/\Gamma$, where $\Gamma = \mathbf{Z} \oplus 2i \mathbf{Z}$. Since $\Gamma$ has index $2$ in $\Lambda$, we have an étale double cover (degree $2$ isogeny) $p \colon X \to Y$.

However, the automorphism $f \colon Y \to Y$ does not lift to $X$. The point is that the lattice $\Gamma$ is not invariant under $f_*$, in fact $f_*(\mathbf{Z} \oplus 2i\mathbf{Z}) = 2\mathbf{Z} \oplus i\mathbf{Z}$