MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

The smallest examples are in dimension $2$. Let $E$ be any elliptic curve over $\mathbf{Q}$ without complex multiplication, e.g., $X_0(11)$. We will construct two twists of $E^6$ E^2$that are isomorphic over$\mathbf{Q}_p$for all$p \le \infty$but not isomorphic over$\mathbf{Q}$. Let$K:=\mathbf{Q}(\sqrt{-1},\sqrt{17})$. Let$G:=\operatorname{Gal}(K/\mathbf{Q})$.G:=\operatorname{Gal}(K/\mathbf{Q}) = (\mathbf{Z}/2\mathbf{Z})^2$. Let $\alpha'$ be the composition of David's homomorphism $\alpha \colon G \to S_6$ with the permutation representation $S_6 \to \operatorname{GL}_6(\mathbf{Z}) operatorname{GL}_2(\mathbf{Z}) = \operatorname{Aut}(E^6)$, operatorname{Aut}(E^2)$be a homomorphism sending the two generators to the reflections in the coordinate axes of$\mathbf{Z}^2$, and let$A$be the$K/k$-twist K/\mathbf{Q}$-twist of $E^6$ E^2$given by$\alpha'$.\alpha$. Define $\beta'$ \beta$and$B$similarly. A matrix calculation shows that no element , but with the lines$y=x$and$y=-x$in place of the coordinate axes. The representations$\operatorname{GL}_6(\mathbf{Z})$conjugates \alpha$ and $\alpha'$ into \beta$of$\beta'$: indeed, the linear conditions G$ on $\mathbf{Z}^2$ are not conjugate: only the coefficients of a conjugating matrix force it to have a determinant that former is 0 mod 2.such that the lattice vectors fixed by nontrivial elements of $G$ generate all of $\mathbf{Z}^2$. Thus $A$ and $B$ are not isomorphic over $\mathbf{Q}$.
On the other hand, every decomposition group $D_p$ in $G$ is smaller than $G$ since $-1$ is a square in $Q_{17}$ and $17$ is a square in $\mathbf{Q}_2$. SoAlso, after restriction to any $D_p$, the homomorphisms restrictions of $\alpha'$ \alpha$and$\beta'$become \beta$ to any proper subgroup of $G$ are conjugatevia some permutation matrix : any single line spanned by a primitive vector in $\operatorname{GL}_6(\mathbf{Z})$.\mathbf{Z}^2$can be mapped to any other by an element of$\operatorname{GL}_2(\mathbf{Z})$. Thus$A$and$B$become isomorphic after base extension to$\mathbf{Q}_p$for any$p \le \infty$.$\square$Remark: The abelian surfaces$A$and$B$constructed above are isogenous even over$\mathbf{Q}$, because the$\mathbf{Z}^2$with one Galois action can be embedded into the$\mathbf{Z}^2$with the other Galois action: rotate$45^\circ$and dilate. Remark: The nonexistence of examples in dimension$1$follows from these two well-known facts: 1) Twists of an elliptic curve over a field$k$of characteristic$0$are classified by$H^1(k,\mu_n)=k^\times/k^{\times n}$where$n$is 2, 4, or 6. 2) If$n<8$, the map$k^\times/k^{\times n} \to \prod_v k_v^\times/k_v^{\times n}$is injective. [Edit: This answer was edited to simplify the construction and to add those remarks at the end.] 1 (If you upvote this answer, please consider upvoting the answers by Felipe Voloch and David Speyer too, since this answer builds on their ideas.) Let$E$be any elliptic curve over$\mathbf{Q}$without complex multiplication, e.g.,$X_0(11)$. We will construct two twists of$E^6$that are isomorphic over$\mathbf{Q}_p$for all$p \le \infty$but not isomorphic over$\mathbf{Q}$. Let$K:=\mathbf{Q}(\sqrt{-1},\sqrt{17})$. Let$G:=\operatorname{Gal}(K/\mathbf{Q})$. Let$\alpha'$be the composition of David's homomorphism$\alpha \colon G \to S_6$with the permutation representation$S_6 \to \operatorname{GL}_6(\mathbf{Z}) = \operatorname{Aut}(E^6)$, and let$A$be the$K/k$-twist of$E^6$given by$\alpha'$. Define$\beta'$and$B$similarly. A matrix calculation shows that no element of$\operatorname{GL}_6(\mathbf{Z})$conjugates$\alpha'$into$\beta'$: indeed, the linear conditions on the coefficients of a conjugating matrix force it to have a determinant that is 0 mod 2. Thus$A$and$B$are not isomorphic over$\mathbf{Q}$. On the other hand, every decomposition group$D_p$in$G$is smaller than$G$since$-1$is a square in$Q_{17}$and$17$is a square in$\mathbf{Q}_2$. So, after restriction to any$D_p$, the homomorphisms$\alpha'$and$\beta'$become conjugate via some permutation matrix in$\operatorname{GL}_6(\mathbf{Z})$. Thus$A$and$B$become isomorphic after base extension to$\mathbf{Q}_p$for any$p \le \infty\$.