Let $k \subset K$ be an extension of algebraically closed fields of characteristic $0$ (e.g. $\overline{\mathbb{Q}} \subset \mathbb{C}$).



> 
> Let X be a smooth variety over $k$. Then is the natural map
>  $$\mathrm{Pic}(X)\,/\,n \to \mathrm{Pic}(X\_K)\,/\,n$$
>  and isomorphism for all $n \in \mathbb{Z}$?
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Here for an abelian group $A$ I denote by $A \, / \, n := A/nA$. Also $X\_K$ denotes the base change of $X$ to $K$.


I can prove this if $X$ is proper using the theory of the Picard scheme. Here we have $\mathrm{Pic}^0(X)\,/\,n = 0$ (the set of points on an abelian variety over an algebraically closed field is divisible). The quotient $\mathrm{Pic}(X)/\mathrm{Pic}^0(X)$ is the Neron-Severi group, which is a finitely generated abelian group scheme hence its points don't change upon base change to $K$.


This proof breaks down for non-proper $X$ as the Picard functor is not representable in general.



Picard group under base change for algebraically closed fields
$\require{AMScd}$
The following result holds in much more generality than asked by the OP. There is no characteristic zero/smoothness assumption.



> 
> 
> > 
> > **Proposition:** Let $X/k$ be a qcqs scheme with $k$ separably closed. Let $K/k$ be a separable extension with $K$ separably closed. Let $n \in \mathbf{N}$ be a positive integer that is prime to the characteristic of $k$. Then the natural map
> >  $$\frac{\operatorname{Pic}(X)}{n\operatorname{Pic}(X)} \to \frac{\operatorname{Pic}(X\_K)}{n\operatorname{Pic}(X\_K)}$$
> >  is an isomorphism.
> > 
> > 
> > 
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The proof goes as follows. The Kummer sequence gives the following commutative diagram with exact rows:
$\require{AMScd}$
\begin{CD}
0 @>{}>>\frac{\operatorname{Pic}(X)}{n\operatorname{Pic}(X)} @>{}>> H^2(X, \mu\_n)@>{}>> H^2(X,\mathbf{G}\_m) \\
{} @VVV @VVV @VVV {} \\
0 @>{}>> \frac{\operatorname{Pic}(X\_K)}{n\operatorname{Pic}(X\_K)} @>{}>> H^2(X\_K, \mu\_n) @>{}>> H^2(X\_K, \mathbf{G}\_m) 
\end{CD}


The middle vertical arrow is an isomorphism by the smooth base change theorem in \'{e}tale cohomology (Corollary 1.3.1 of [this](http://math.stanford.edu/~conrad/Weil2seminar/Notes/L7-8.pdf)):



> 
> 
> > 
> > **Smooth Base Change:** Suppose $K/k$ is an extension of separably closed fields, $X$ a qcqs $k$-scheme and $\mathscr{F}$ a torsion sheaf with torsion orders not divisible by the characteristic of $k$.Then, the base change map
> >  $$H^i(X,\mathscr{F}) \to H^i(X\_K, \mathscr{F}\_K)$$
> >  is an isomorphism for all $i \geq 0$.
> > 
> > 
> > 
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Therefore to prove the proposition, we reduce to proving the following injectivity statement concerning Brauer groups:



> 
> 
> > 
> > **Proposition:** Let $k \subseteq K$ be a separable extension of fields with $k$ separably closed. Let $X/k$ be a qcqs scheme. Then $H^2(X,\mathbf{G}\_m)$ injects into $H^2(X\_K, \mathbf{G}\_m)$. 
> > 
> > 
> > 
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To prove this, write $K$ as a direct limit of smooth $k$-algebras. Since cohomology commutes with direct limits for qcqs schemes, if a (cohomological) Brauer class $\alpha$ in $H^2(X,\mathbf{G}\_m)$ dies in $H^2(X\_K,\mathbf{G}\_m)$ it must die in $H^2(X\_R, \mathbf{G}\_m)$ for some smooth $k$-algebra $R$. By smoothness and using that $k$ is separably closed, $\operatorname{Spec} R$ has a $k$-rational point. Therefore the canonical map $f: X\_{R} \to X$ has a section $g$ and so on cohomology the composition
$$H^2(X,\mathbf{G}\_m) \stackrel{f^\ast}{\to} H^2(X\_R,\mathbf{G}\_m) \stackrel{g^\ast}{\to} H^2(X,\mathbf{G}\_m)$$ is the identity. We conclude that $\alpha = 0$ as desired.


Finally we remark that in the injectivity statement on Brauer groups, only the ground field is required to be separably closed. There is no requirement on $K$, only that $K/k$ be separable.


**Edit:** The proof also shows that for any \'{e}tale sheaf $\mathscr{F}$ on $X$ such that multiplication by $n$ is surjectve, then $$\frac{H^i(X,\mathscr{F})}{nH^i(X,\mathscr{F})} \to \frac{H^i(X\_K,\mathscr{F})}{nH^i(X\_K,\mathscr{F})} $$
is an isomorphism for all $i \geq 1$.


