Are "large enough" finite etale covers arithmetic? Let $X$ be a variety over a number field $K$. Then it is known that for any topological covering $X' \to X(\mathbb{C})$, the topological space $X'$ can be given the structure of a $\overline{K}$-variety in such a way so that the morphism $f: X' \to X$ inducing the topological map is a finite etale morphism over $\overline{K}$. However, the variety $X'$ and the morphism $f$ may not descend to $K$.
My question is as follows: does there always exist a further finite etale covering $f' : X'' \to X'$ such that the composition $X'' \to X$ may be defined over $K$?
EDIT: Just to be clear, I'd like all the covers involved to be geometrically connected to avoid trivial solutions.
 A: Here's a simple argument assuming $X$ admits a $K$-rational point, and that $X$ has a finitely generated geometric fundamental group. In fact the "further" covering $X''$ can be chosen to be geometrically Galois over $X$.
Let $\Pi := \pi_1(X_K)$, let $\overline{\Pi} := \pi_1(X_{\overline{K}})$ (assumed to be topologically finitely generated). Let $G_K := \text{Gal}(\overline{K}/K)$.
Since we're working over a field, there's a homotopy exact sequence
$$1\rightarrow \overline{\Pi}\rightarrow\Pi\rightarrow G_K\rightarrow 1$$
from which we get a canonical outer action $G_K\rightarrow\text{Out}(\overline{\Pi})$.
The covering $X'$ (over $\overline{K})$ corresponds to a finite index subgroup $H \le \overline{\Pi}$. It would suffice to find a finite index normal subgroup $\Gamma\lhd \overline{\Pi}$ which is stabilized by $G_K$. Indeed, using the $K$-rational point of $X$, the homotopy exact sequence is split, so the outer action of $G_K$ comes from an honest action, and $\Pi = \overline{\Pi}\rtimes G_K$ relative to this action. If $\Gamma\lhd\overline{\Pi}$ is stabilized by $G_K$, then the subgroup $\Gamma\rtimes G_K\le \Pi$ visibly corresponds to a geometrically connected finite cover of $X_K$ (though it may not be normal inside $\Pi$).
To find this $\Gamma$, let $N\le H$ be the intersection of all the $\overline{\Pi}$-conjugates of $H$, so $N$ is normal and of finite index inside $\overline{\Pi}$. Let $\Gamma$ be the intersection of the kernels of all the surjective homomorphisms $\overline{\Pi}\rightarrow\overline{\Pi}/N$. Since $\overline{\Pi}$ is finitely generated, there are only finitely many such homomorphisms, so $\Gamma$ is also finite index inside $\overline{\Pi}$. Moreover, it's easy to check that $\Gamma$ is characteristic inside $\overline{\Pi}$. Thus, $G_K$ must stabilize $\Gamma$, and hence $\Gamma\rtimes G_K$ will correspond to the desired covering $X_K''\rightarrow X_K$, which is moreover geometrically Galois.
A: Adding on Will's and Sasha's answers, the condition of having a rational point, or at least a "1-truncated homotopy fixed point" for the action is necessary. For example, let $C_2$ act on the circle $S^1$ by half rotation. The covers of $S^1$ are the standard n-fold ones, and we can ask what it takes to lift the action of $C_2$ to the cover, so that it is "defined over $BC_2$". In particular, we need to lift that half-circle rotation to the n-fold cover, for which the options are $1/2n + k/n$ rounds rotation. For this to be an involution, we need that applying it twice gives the identity, i.e. that $1/n +2k/n$ is an integer. If $n$ is even, this is impossible, and so the double cover of this action on $S^1$ has no cover definable over $BC_2$. To turn this topological picture to arithmetic, take $K=\mathbb{R}$ and let complex conjugation act on $\mathbb{C}^\times$ by $z\mapsto -1/\bar{z}$ (which is a form of the multiplicative group with no rational points). The action on the unit circle is then half rotation, so the Galois story realized to the topological one up to profinite completion.
I would add that what happens topologically is that if we have a fixed point, we can use it to define a "connected" compositum of pointed covers, by taking the component of the tuple of base points lifts. This is what missing in this example esssentially, even though up to isomorphism all covers are actually "the same".
A: Let's assume that $X$ admits a $K$-point $x$ and use the corresponding geometric point as the base point. The existence of a rational point is in fact necessary for a positive answer, as explained by S. carmeli.
In terms of etale fundamental groups the question can be paraphrased as follows: given an open subgroup $H\subset \pi_1(X_{\overline{K}},x)$ does there exist an open subgroup $H'\subset H$ such that the action of the Galois group $G_K$ on $\pi_1(X_{\overline{K}},x)$ preserves $H'$.
This is true and follows from $\pi_1(X_{\overline{K}},x)$ being topologically finitely generated. Consider the subgroup $\Gamma_H\subset G_K$ consisting of elements $\gamma\in G_K$ such that $\gamma(H)=H$. Let $h_1,\dots, h_n$ be a set of topological generators of $H$ ($H$ is topologically finitely generated because it has finite index in $\pi_1(X_{\overline{K}})$). Then $\Gamma_H$ can be expressed as $\{\gamma\in G_K|\gamma(h_i)\in H\}$ so $\Gamma_H$ is an intersection of finitely many open subset, hence is an open subgroup. In particular, $\Gamma_H$ has finite index in $G_K$. Take $\Gamma\subset \Gamma_H$ to be an open subgroup which is moreover normal in $G_K$.
Let $g_1,\dots, g_m$ be a set of representatives of cosets of $\Gamma$ in $G_K$. Then $H'=\bigcap g_i(H)$ is an open subgroup with the desired property. Indeed, suppose that $x\in H'$ and $\gamma g_i\in G_K$ are arbitrary elements where $\gamma\in \Gamma$ and $i\in\{1,\dots, m\}$. The result of the action $\gamma \circ g_i(x)$ lies in $H'$ because for each $k=1,\dots, m$ we have $g_k^{-1}\gamma g_i=\gamma'g_j^{-1}$ for some $\gamma'\in \Gamma$ and $j\in\{1,\dots, m\}$ so $\gamma g_i(x)\in \gamma g_ig_j(H)=g_k\gamma'(H)=g_k(H)$.
We can think of this argument as of a generalization of the proof that a compact group acting on a finite-dimensional $\mathbb{Q}_p$-vector space always preserves some $\mathbb{Z}_p$-lattice.
