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Let $k$ be a number field, and let $X$ be a projective $k$-variety. Let $mathcalV$ be a vector bundle on $X$ that is defined over $k$. A vector bundle $mathcalV'$ on $X$ that is defined over $k$ is said to be a twist of $mathcalV$ if there is some finite extension $ell/k$ such that we have an isomorphism $mathcalV_ell simeq mathcalV'_ell$ of bundles upon basechanging to $ell$.

Question: Is it true that all twists of $mathcalV$ are in fact isomorphic to $mathcalV$ over $k$?

What I know: Twists are parameterized by the Galois cohomology group $H^1(operatornameGal(overlinek/k),operatornameAut(calV_overlinek))$, so all we need to do is determine whether this cohomology group is equal to $0$. It's a consequence of Hilbert Theorem 90 that $H^1(operatornameGal(overlinek/k),operatornameGL_n(overlinek)) = 0$, which might be useful.

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Thanks to a very helpful conversation with my friend P. Achinger here is a solution assuming that $X$ is geometrically integral (he has a proof even when $X$ is just proper and $k$ is infinite, but below is a simplified version in the case when geometrically integral case when $k$ is infinite).

Assume that $k$ is infinite.

Step 1: The assertion is true for line bundles. Indeed, if $L_ovkcong L'_ovk$ then $(Lotimes L^-1)_ovkcong (h_X)_ovk$. But, this then defines a a class of $H^1(Gal(ovk/k),Aut(h_X_ovk))=H^1(Gal(ovk/k),ovk^times)=0$. It follows that $Lcong L'$.

Step 2: Let $V$ and $V'$ be vector bundles on $X$ such that $V_ovkcong V'_ovk$. Note then that $det(V)_ovkcong det(V')_ovk$ so that, by Step 1, we have that $det(V)cong det(V')$. Let us now consider the functor $Rmapsto Hom(V_R,V'_R)$. Note that by flat base change we have that this functor coincides with $Rmapsto Hom(V,V')otimes_k R$ and thus coincides with $A^n_k$ where $n:=dim Hom(V,V')$ (which is finite-dimensional since $X$ is proper). Let $Usubseteq A^n_k$ be defined as the functor
$$U(R)=finHom(V_R,V'_R):ftext isomorphism$$

We claim that this is an open subscheme of $A^n_k$. Indeed, note that we have a functorial map

$$beginalignedA^n_k =&Hom(V,V')toHom(det(V),det(V))=A^1_k\ &:varphimapstodet(varphi)endaligned$$

(Where we've used the fact that $X$ is geometrically integral to identify this last term with $A^1_k$). It's clear then that $U$ is the open subscheme of $A^n_k$ obtained as the preimage of

$$mathbbG_msubseteq A^1_k=Hom(det(V),det(V))$$

Now by assumption we have that $U(ovk)ne varnothing$ so that $U$ is non-empty. This implies, since $k$ is infinite, that $U(k)ne varnothing$ (e.g see this). Thus, $Vcong V'$.

EDIT: Here's a proof that works when $k$ is finite, and thus all cases are covered.

Let us denote by $G_V$ the group scheme over $k$ given by $Rmapsto Aut(V_R)$. Note that, as above, we have that $G_V$ is an open subscheme of the functor $Rmapsto mathrmEnd(V_R)$ which is identified with $A^n_k$ where $n:=dimmathrmEnd(V)$. In particular, since $A^n_k$ is irreducible we deduce that $G_V$ is connected.

Note then that since the twists of $V$ are classified by $H^1_mathrmet(mathrmSpec(k),G_V)$ it suffices to prove that this latter cohomology group is zero. Since $G_V$ is connected this follows from Lang's Theorem.

Remark: It's still a little miraculous that even though the group $G_V:Rmapsto mathrmAut(V_R)$ is mysterious (at least to me) we can prove that $H^1_mathrmet(mathrmSpec(k),G_V)=0$. Does anyone have anything substantive to say about the structure of the group $G_V$? Is it affine? Is it reductive (note that it can't be semisimple because we have a central embedding of $mathbbG_m$ into $G_V$)? If so, is it split? For example, what made the line bundle case so simple is that for any line bundle $L$ we have that $G_LcongmathbbG_m$. Unless I made a mistake, it seems like that it needn't be reductive in general since some examples on $mathbbP^1$ lead me to believe that you can get things like parabolic groups in $GL_n$.