Galois invariant Picard group elements 
Let $X$ be a smooth variety over a perfect field $k$ with $X(k) \neq \emptyset$. Then is the natural map
  \begin{equation}
 \mathrm{Pic}(X) \to  (\mathrm{Pic}(X_{\bar{k}}))^{\mathrm{Gal}(\bar{k}/k)} \qquad (1)
\end{equation}
  surjective?

Remarks:


*

*If $X$ is a proper, then the map (1) is in fact an isomorphism. This is usually proved using the Hochshild--Serre spectral sequence.

*The map (1) need not be injective in general. Take $X$ to be the complement of a closed point of degree two in  $\mathbb{P}^1_k$. Then $\mathrm{Pic}(X) = \mathbb{Z}/2\mathbb{Z}$ but $\mathrm{Pic}(X_{\bar{k}}) = 0$.

 A: Updated. The example by @Lucifer is completely correct.  Thanks to @Count Dracula who explained the example proposed by @Lucifer. 
That example is fine.  I am keeping the counterexamples below, since they arise in a different way: as Severi-Brauer schemes over multiplicative group schemes over a field (and schemes mapping to such Severi-Brauer schemes).  The counterexamples are given after a general statement that can (sometimes) be used to prove surjectivity of the morphism (1).  
Setup. Denote by $S$ the scheme $\text{Spec}\ k$ (to save typing).  As in Borovoi-- Colliot-Thélène -- Skorobogatov, denote by $\mathfrak{g}$ the Galois group $\text{Gal}(\overline{k}/k)$.  Let $\pi:X\to S$ be a smooth, separated, quasi-compact morphism with geometrically irreducible fiber.  Let $\sigma:S\to X$ be a section of $\pi$.  Denote by $G_X$, resp. $G_{X,\sigma}$, the group $S$-scheme that represents the functor sending every smooth $S$-scheme $T$ to $\mathbb{G}_{m,X}(X\times_S T)$, resp. to the kernel of $$\sigma^*(T):\mathbb{G}_{m,X}(X\times_S T) \to \mathbb{G}_{m,S}(T).$$  Pullback by $\sigma$ defines a group homomorphism, $$\sigma^*:\text{Br}(X)\to \text{Br}(S).$$  Denote the kernel by $\text{Br}(X)_\sigma$. Pullback by $\pi$ defines a group homomorphism, $$\pi^*:H^2_{\mathfrak{g}}(G_{X,\sigma}(\overline{k})) \to \text{Br}(X)_\sigma.$$
Lemma. The cokernel of the natural restriction morphism (1) equals the kernel of $H^2_{\mathfrak{g}}(G_{X,\sigma}(\overline{k})) \to \text{Br}(X)_\sigma$.  The group $G_{X,\sigma}(\overline{k})$ is a finitely generated, torsion-free Abelian group.  In particular, the cokernel of the natural restriction morphism is a torsion group.
Proof. The relevant terms in the Hochschild-Serre spectral sequence give the following long exact sequence, $$0 \to H^1_{\mathfrak{g}}(G_X(\overline{k})) \to \text{Pic}(X)\xrightarrow{\text{res}} \text{Pic}(X_{\overline{k}})^{\mathfrak{g}} \xrightarrow{\delta} H^2_{\mathfrak{g}}(G_X(\overline{k})) \to \text{Br}(X).$$  By Hilbert's Theorem 90, the first term is $H^1_{\mathfrak{g}}(G_{X,\sigma}(\overline{k}))$.  Pullback by the section $\sigma$ defines a splitting of the map $$H^2_{\mathfrak{g}}(\mathbb{G}_{m,S}(\overline{k}))\to \text{Br}(X).$$  Thus, the cokernel of $\text{res}$ equals the kernel of $\pi^*$.
The group $G_{X,\sigma}(\overline{k})$ is finitely generated by Rosenlicht.  Every torsion-element satisfies an integral equation over $\overline{k}$.  Since $X_{\overline{k}}$ is integral, $\overline{k}$ is integrally closed in $H^0(X_{\overline{k}},\mathcal{O}_X)$.  Thus, $G_{X,\sigma}(\overline{k})$ is a finitely generated, torsion-free Abelian group.  The profinite Galois cohomology group $H^2_{\mathfrak{g}}(G_\sigma(\overline{k}))$ is torsion.  Thus, $H^2_{\mathfrak{g}}(G_\sigma(\overline{k}))$ is a torsion group.  QED
The contravariant functor $G$ that sends a pointed $S$-scheme, $(X,\sigma)$, to the group scheme $G_{X,\sigma}$ has an adjoint.  For every smooth group $S$-scheme $M$ with $M\times_{\text{Spec}\ k} \text{Spec}\ \overline{k}$ isomorphic to $\mathbb{G}_{m,\overline{k}}^\rho$, the group scheme $G_{M,e}$ is the Cartier dual group scheme $M^D$.  For every morphism of group $S$-schemes, $$\phi:M^D\to G_{X,\sigma},$$ there is a unique $S$-morphism, $$f:X\to M,$$ sending $\sigma$ to the identity section $e$ and such that the pullback map on unit group $S$-schemes is $\phi$.
Proposition. The Galois cohomology group $H^2_{\mathfrak{g}}(M^D)$ is canonically isomorphic to the kernel, $\text{Br}_1(M)$, of the pullback map $$\text{pr}_M^*:\text{Br}(M)\to \text{Br}(M\times_{\text{Spec}\ k} \text{Spec}\ \overline{k}).$$  
Proof. Since $\text{Pic}(\mathbb{G}_{m,\overline{k}}^\rho)$ is the trivial group, this follows from the previous lemma. QED
The Cartier dual group $S$-scheme to $G_\sigma$ is a smooth group $S$-scheme $M$ of multiplicative type. There is a unique $S$-morphism $$f:X\to M,$$ mapping $\sigma$ to the identity section $e$ and such that the induced homomorphism of groups of units is an isomorphism of $S$-group schemes.  
Proposition. The cokernel of $\text{res}$ is identified with the kernel of the group homomorphism $$f^*:\text{Br}_1(M) \to \text{Br}(X).$$
In particular, if there exists a degree $1$ zero-cycle in the generic fiber of $f$, then $\text{res}$ is surjective.
Proof. This follows from the previous proposition and functoriality of the Hochschild-Serre spectral sequence with respect to $f$.  If the fiber of $f$ over $\eta=\text{Spec}\ k(M)$ has a degree $1$ zero-cycle, then the pullback map $\text{Br}(k(M))\to \text{Br}(X_\eta)$ is injective by the lemma.  Since $M$ is smooth, it follows that also $f^*$ is injective.   QED
Counterexamples. This suggests how to construct a counterexample, namely as a Severi-Brauer scheme over a multiplicative group scheme $M$ over $k$.  
Let $k$ be a finite field $\mathbb{F}_q$.  Let $M$ be $\mathbb{G}_{m,k}= \text{Spec}\ k[t,t^{-1}]$.  The Cartier dual group scheme $M^D$ has $M^D(k) = M^D(\overline{k}) \cong \mathbb{Z}$, generated by the element $t$.  
For every integer $\ell$, let $k_\ell=\mathbb{F}_{q^\ell}$ be a degree $\ell$ extension of $k$ with cyclic Galois group generated by a Frobenius map.  This extension together with the generator of the Galois group is unique up to unique isomorphism.  Associated to the cyclic extension $k_\ell/k$ and the element $t\in M^D(k)$, there is a cyclic algebra $A_{\ell,t}$ that is a locally free $\mathcal{O}_M$-module of rank $\ell^2$ and that is an Azumaya algebra over $\mathcal{O}_M$.  The Brauer class of $A_{\ell,t}$ in $\text{Br}(M)$ is an element of order $\ell$ generating the full $\ell$-torsion subgroup of $\text{Br}(M)$.  By Tsen's Theorem, the Brauer group of $M\times_{\text{Spec}\ k} \text{Spec}\ \overline{k}$ is trivial.  Thus, the class of $A_{\ell,t}$ generates a cyclic subgroup of order $\ell$ in $\text{Br}(M)_k$.  
Now let $f:X\to M$ be the Severi-Brauer scheme parameterizing left ideals in $A_{\ell,t}$ of rank $\ell$.  Since the Brauer group of $k$ is trivial (by Wedderburn or Chevalley-Warning), the restriction of $f$ over the $k$-rational point $e\in M(k)$ is a trivial Severi-Brauer scheme.  Thus, there exists a section $$\sigma:\text{Spec}\ k \to X,$$ with image in the fiber of $f$ over $e$.  The Picard group of $X\times_{\text{Spec}\ k} \text{Spec}\ \overline{k}$ is $\mathbb{Z}$, and the generator restricts on each projective space (geometric) fiber of $f$ to a generator $\mathcal{O}(1)$ of the projective space.  However, there cannot be such an invertible sheaf on $X$, or else $\mathcal{A}_{\ell,t}$ would have zero Brauer class.    
A: This is not true. Let $k$ be a field of characteristic not $2$ and $E$ is an elliptic curve with origin $0$ and $k$-rational $2$-torsion point $P$. Assume $E$ has a third rational point. Then $X = E \setminus \{0, P\}$ has at least one rational point. We have
$$
Pic(X) = E(k)/\langle P \rangle
$$
But the Galois invariants in $Pic(X_{\overline{k}})$ are equal to $E'(k)$ where $E \to E'$ is the isogeny whose kernel is the order $2$ subgroup scheme generated by $P$. Thus all you have to do is find an example of $k, E, P$ where $E(k) \to E'(k)$ is not surjective. For example if the ground field $k$ is finite this will be the case because $|E(k)| = |E'(k)|$ in that case.
