I can prove $(3) \Rightarrow (1)$ when $X$ and $Y$ are irreducible (Lemma 1 below; no integrality assumptions needed. Irreducible is needed for the question to make sense), and I can prove $(4) \Rightarrow (1)$ when $X$ and $Y$ are normal, connected, and Noetherian (in particular integral, see Tag 033M) (Lemma 2 below).
I do not know how to approach the general case if $Y$ is not normal; see the remark at the end. Maybe you can weaken it to something like unibranch? The property that I need is that a connected finite étale cover of $Y$ is always integral; I don't know in what generality this is true.
Reference. A good introductory reference to some of the stuff I'm using is the chapter Fundamental groups of schemes of the Stacks project. (I make precise references as I go along.)
Notation. Let $\eta_X$ (resp. $\eta_Y$) be the generic point of $X$ (resp. $Y$). I will use the following definition.
Definition. Let $\phi \colon X \to Y$ be a morphism of schemes, let $G$ be a finite group (viewed as constant étale group scheme), and let $G \times X \to X$ be an action of $Y$-schemes. Then $\phi$ is Galois with group $G$ if the induced map $G \times X \to X \times_Y X$ is an isomorphism.
If $F \colon \operatorname{FÉt}_Y \to \operatorname{Set}$ is a fibre functor (notably, $F = F_{\bar y}$ associated to taking the geometric fibre over a geometric point $\{\bar y\} \to Y$), there is the following equivalent notion of being Galois.
Auxiliary lemma. Let $\phi \colon X \to Y$ be a finite étale morphism of connected schemes. Then $|\!\operatorname{Aut}_Y(X)| \leq |F(X)|$, and equality holds if and only if $\phi$ is Galois.
Proof. Let $G = \operatorname{Aut}_Y(X)$, with the natural action on $X$, and consider the induced map
$$\psi \colon G \times X \to X \times_Y X.$$
Applying the fibre functor $F = F_{\bar y}$ gives
\begin{align*}
F(\psi) \colon G \times F(X) &\to F(X) \times F(X)\\
(\sigma, s) &\mapsto (\sigma(s),s).
\end{align*}
By Tag 0BN0 (7), this map is injective, and it is bijective if and only if $|\!\operatorname{Aut}_Y(X)| = |F(X)|$.
Thus we need to show that $\psi$ is an isomorphism if and only if $F(\psi)$ is a bijection. This follows since $F$ reflects isomorphisms (see Tag 0BMY and Tag 0BNB). $\square$
Lemma 1. Let $\phi \colon X \to Y$ be a finite étale morphism of irreducible schemes. Assume there exists a finite group $G$ and an action $G \times X \to X$ whose base change to $\{\eta_X\}$ makes $\{\eta_X\} \to \{\eta_Y\}$ a Galois extension with group $G$. Then $X \to Y$ is Galois with group $G$.
Proof. By the auxiliary lemma, we only have to show that $X$ has enough automorphisms. This is now a simple degree/counting argument (the elements of $G$ providing the automorphisms). $\square$
Lemma 2. Let $\phi \colon X \to Y$ be a finite étale morphism of connected normal Noetherian schemes, and assume that the extension $\{\eta_X\} \to \{\eta_Y\}$ is Galois. Then $\phi$ is Galois.
Proof. We want to apply Tag 0BN6 $(3) \Rightarrow (2)$ to the pullback functor
\begin{align*}
H \colon \operatorname{FÉt}_Y &\to \operatorname{FÉt}_{\eta_Y}\\
Z &\mapsto Z_{\eta_Y}.
\end{align*}
Assume $Z$ is a connected scheme finite étale over $Y$. Then $Z$ is normal (Tag 025P), connected, and Noetherian; hence integral (Tag 033M). Hence, it only has one generic point $\eta_Z$, and this is the only point lying over $\eta_Y$. Thus, $H(Z)$ is connected, so Tag 0BN6 $(3) \Rightarrow (2)$ implies that $H$ is fully faithful.
Note that $H$ also reflects isomorphisms, since $F_{\overline{\eta_Y}}$ does, which factors through $H$. For a fully faithful functor that reflects isomorphisms, we have
$$\operatorname{Aut}_Y(X) = \operatorname{Aut}_{H(Y)}(H(X)),$$
so $X$ has enough automorphisms since $H(X)$ does so. $\square$
Remark. We see that $H$ is fully faithful if and only if for every connected étale $Y$-scheme $Z$, the generic fibre $Z_{\eta_Y}$ is just a point. This is certainly the case if $Z$ is integral, and I think the converse holds as well. (See below for an example where $Z$ has multiple components.)
Also, we might not need the full strength of $H$ being fully faithful: we only need to get back all automorphisms.
Example. Let $A = \mathbb R[x,y]/(x^2+y^2)$. Then $A$ is a domain. It has a finite étale extension
$$B = \mathbb C[x,y]/(x^2+y^2) = \mathbb C[x,y]/(x+iy) \times \mathbb C[x,y]/(x-iy),$$
which is connected but not integral. The fibre above the generic point consists of two points. Correspondingly, if $Y = \operatorname{Spec} A$, the functor
$$H \colon \operatorname{FÉt}_Y \to \operatorname{FÉt}_{\eta_Y}$$
is not fully faithful, by Tag 0BN6 again. I do not know whether there exists an integral finite étale $A$-algebra $B$ for which $\operatorname{Frac}(A) \to \operatorname{Frac}(B)$ is Galois, but $A \to B$ is not.
