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Are there counterexamples to the following: Given two varieties $A$, $\tilde{A}$, both defined over $\mathbb{Q}$, one of which, say $A$, is a Shimura variety. Then, every isomorphism defined over $\overline{\mathbb{Q}}$ $$f: A_{\overline{\mathbb{Q}}}\rightarrow \tilde{A}_{\overline{\mathbb{Q}}}$$ can be twisted to an isomorphism $$f^{*}: A_{{\mathbb{Q}}^{\text{solv}}}\rightarrow \tilde{A}_{\mathbb{Q}^{\text{solv}}}$$ defined over ${\mathbb{Q}}^{\text{solv}}$, the solvable closure of $\mathbb{Q}$.

If one could show that this holds at least for genus-one curves, this would imply a long sought-after conjecture that every genus-one curve over $\mathbb{Q}$ has a solvable point (every genus-one curve is isomorphic over $\overline{\mathbb{Q}}$ to its Jacobian, which is an elliptic curve over $\mathbb{Q}$ and has solvable points), so I want to ask: Are any results known about this in higher genus?

To my knowledge, the best result to this day is that of Ciperiani and Wiles (https://www.ma.utexas.edu/users/mirela/solvable.pdf), who give an affirmative answer in the case that the genus-one curve is locally nontrivial and semistable, but neither of these conditions seems to be necessary.

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In good cases, the ''isomorphism functor'' $\mathrm{Isom}(X,Y)$ is representable by a scheme. Hence, you are asking that if this scheme is non-empty (i.e. contains a geometric point), whether it contains a solvable point. So your problem is more-or-less equivalent to the usual open problem that every variety contains a solvable point.

Now for your special cases. If $X$ and $Y$ are curves of genus $1$, then $\mathrm{Isom}(X,Y)$ is a torsor under $\mathrm{Aut}(X)$. In particular, to find a solvable point on it, one quickly reduces to finding a solvable point on some torsor of the Jacobian of $X$, i.e. some curve of genus $1$, which equivalent to the original problem.

For curves of higher genus, the scheme $\mathrm{Aut}(X)$ will be finite, so you reduce to the easier problem of asking whether this is a solvable group (you should be able to find counter-examples to your question here, if e.g. $\mathrm{Aut}(X_{\bar{\mathbb{Q}}})$ is a simple non-solvable group, which occurs if $X$ is the Klein quartic, say). Moreover here, in general, $X$ won't have a twist that contains a rational point (e.g. this happens if $X(\mathbb{Q}) = \emptyset$ and $\mathrm{Aut}(X)=\{\mathrm{id}\}$). So it seems that your approach to solving this conjecture is also doomed to fail here.

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