Nontrivial Twists of a Vector Bundle Let $k$ be a number field, and let $X$ be a projective $k$-variety. Let $\mathcal{V}$ be a vector bundle on $X$ that is defined over $k$. A vector bundle $\mathcal{V}'$ on $X$ that is defined over $k$ is said to be a twist of $\mathcal{V}$ if there is some finite extension $\ell/k$ such that we have an isomorphism $\mathcal{V}_\ell \simeq \mathcal{V}'_\ell$ of bundles upon basechanging to $\ell$.
Question: Is it true that all twists of $\mathcal{V}$ are in fact isomorphic to $\mathcal{V}$ over $k$?
What I know: Twists are parameterized by the Galois cohomology group $H^1(\operatorname{Gal}(\overline{k}/k),\operatorname{Aut}(\cal{V}_{\overline{k}}))$, 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 $H^1(\operatorname{Gal}(\overline{k}/k),\operatorname{GL}_n(\overline{k})) = 0$, which might be useful.
 A: $\newcommand{\GL}{\mathrm{GL}}$$\newcommand{\ov}[1]{\overline{#1}}$$\newcommand{\h}{\mathcal{O}}$$\newcommand{\Aut}{\mathrm{Aut}}$$\newcommand{\Gal}{\mathrm{Gal}}$$\newcommand{\Hom}{\mathrm{Hom}}$$\newcommand{\A}{\mathbb{A}}$
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 $L_{\ov{k}}\cong L'_{\ov{k}}$ then $(L\otimes L^{-1})_{\ov{k}}\cong (\h_X)_{\ov{k}}$. But, this then defines a a class of $H^1(\Gal(\ov{k}/k),\Aut(\h_{X_{\ov{k}}}))=H^1(\Gal(\ov{k}/k),\ov{k}^\times)=0$. It follows that $L\cong L'$.
Step 2: Let $V$ and $V'$ be vector bundles on $X$ such that $V_{\ov{k}}\cong V'_{\ov{k}}$. Note then that $\det(V)_{\ov{k}}\cong \det(V')_{\ov{k}}$ so that, by Step 1, we have that $\det(V)\cong \det(V')$. Let us now consider the functor $R\mapsto \Hom(V_R,V'_R)$. Note that by flat base change we have that this functor coincides with $R\mapsto \Hom(V,V')\otimes_k R$ and thus coincides with $\A^n_k$ where $n:=\dim \Hom(V,V')$ (which is finite-dimensional since $X$ is proper). Let $U\subseteq \A^n_k$ be defined as the functor 
$$U(R)=\{f\in\Hom(V_R,V'_R):f\text{ isomorphism}\}$$
We claim that this is an open subscheme of $\A^n_k$. Indeed, note that we have a functorial map 
$$\begin{aligned}\A^n_k =&\Hom(V,V')\to\Hom(\det(V),\det(V))=\A^1_k\\ &:\varphi\mapsto\det(\varphi)\end{aligned}$$
(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 
$$\mathbb{G}_m\subseteq \A^1_k=\Hom(\det(V),\det(V))$$
Now by assumption we have that $U(\ov{k})\ne \varnothing$ so that $U$ is non-empty. This implies, since $k$ is infinite, that $U(k)\ne \varnothing$ (e.g see this). Thus, $V\cong V'$. 
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 $R\mapsto \Aut(V_R)$. Note that, as above, we have that $G_V$ is an open subscheme of the functor $R\mapsto \mathrm{End}(V_R)$ which is identified with $\A^n_k$ where $n:=\dim\mathrm{End}(V)$. 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 $H^1_\mathrm{et}(\mathrm{Spec}(k),G_V)$ 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 $G_V:R\mapsto \mathrm{Aut}(V_R)$ is mysterious (at least to me) we can prove that $H^1_\mathrm{et}(\mathrm{Spec}(k),G_V)=0$. 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 $\mathbb{G}_m$ 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 $G_L\cong\mathbb{G}_m$. Unless I made a mistake, it seems like that it needn't be reductive in general since some examples on $\mathbb{P}^1$ lead me to believe that you can get things like parabolic groups in $\GL_n$.
EDIT: Just to point out, the group $G_V$ is evidently affine. Namely, by the discussion above we see that $\Aut(G)$ is just $D(\det)$ or, more precisely, the pullback of $\mathbb{G}_m$ along the map $\mathrm{End}(V)\to\mathrm{End}(\det(V))$ which, since this map is between affine schemes and $\mathbb{G}_m$ is affine, is affine.
