I change the notations!
If I am not wrong: $E$ is a principal $GL(n,\mathbb{R})$-bundle over a topological space $X$, $Ad(E)$ is the adjoint vector bundle (over $X$) associated to $E$ and $F(E)\equiv F$ is the frame (vector) bundle (over $X$) associated to $E$.
By definition: there exists an open covering $\{U_{\alpha}\}_{\alpha\in A}$ (for $Ad(E)$) of $X$ such that:
$\pi_1^{-1}(U_{\alpha})\stackrel{\varphi_{\alpha}}{\cong}U_{\alpha}\times\mathfrak{gl}(n,\mathbb{R})=U_{\alpha}\times\mathbb{R}^n_n$, $\varphi_{\alpha}$ is a homeomorphism;
$pr_1\circ\varphi_{\alpha}=\pi_1$ ;
getting $U_{\alpha\beta}=U_{\alpha}\cap U_{\beta}\neq\emptyset$, the maps:
\begin{equation}
\varphi_{\beta\displaystyle|\pi_1^{-1}(U_{\alpha\beta})}\circ\varphi^{-1}_{\alpha\displaystyle|\pi_1^{-1}(U_{\alpha\beta})}:(P,M)\in U_{\alpha\beta}\times\mathbb{R}^n_n\to(P,Ad(g_{\alpha\beta}(P)^{-1})(M))\in U_{\alpha\beta}\times\mathbb{R}^n_n
\end{equation}
are homeomorphism and the functions $g_{\alpha\beta}:U_{\alpha\beta}\to GL(n,\mathbb{R})$ are the transition functions of $E$.
In the same way: there exists an open covering $\{V_{\alpha}\}_{\alpha\in A}$ (for $End(F)$) of $X$ such that:
$\pi_2^{-1}(V_{\alpha})\stackrel{\psi_{\alpha}}{\cong}V_{\alpha}\times End(\mathbb{R}^n)=V_{\alpha}\times\mathbb{R}^n_n$, $\psi_{\alpha}$ is a homeomorphism;
$pr_1\circ\psi_{\alpha}=\pi_2$ ;
getting $V_{\alpha\beta}=V_{\alpha}\cap V_{\beta}\neq\emptyset$, the maps:
\begin{equation}
\psi_{\beta\displaystyle|\pi_2^{-1}(V_{\alpha\beta})}\circ\psi^{-1}_{\alpha\displaystyle|\pi_2^{-1}(V_{\alpha\beta})}:(P,M)\in V_{\alpha\beta}\times\mathbb{R}^n_n\to\left(P,\left(^Tg_{\alpha\beta}^{-1}\otimes g_{\alpha\beta}\right)(P)(M)\right)\in V_{\alpha\beta}\times\mathbb{R}^n_n
\end{equation}
are homeomorphism and the functions $g_{\alpha\beta}:V_{\alpha\beta}\to GL(n,\mathbb{R})$ are the transition functions of $E$ (and of $F$).
Remark. For any pair of vector bundles $V$ and $W$ over $X$: $Hom(V,W)\cong V^{\vee}\otimes W$!, where $V^{\vee}$ is the dual (vector) bundle of $V$.
Whitout loss of generality, we can assume that $Ad(E)$ and $F$ have the same open covering of trivialization $\{U_i\}_{i\in I}$ over $X$!, by a simply computation:
\begin{gather}
\forall i,j\in I,P\in U_{ij}\neq\emptyset,M\in\mathbb{R}^n_n,\\
Ad(g_{ij}^{-1}(P))(M)=g_{ij}^{-1}(P)\times M\times g_{ij}(P)=\left(^Tg_{ij}^{-1}\otimes g_{ij}\right)(P)(M);
\end{gather}
in other words, because $Ad(E)$ and $End(F)$ have the same open covering of trivialization and the same transition functions, they are canonically isomorphic!
$\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.
Best Answer
One example is the Möbius strip, which admits a homogeneous $E(2)$ (or $SE(2)$) structure. To see this note that the set of affine lines in $\mathbb{R}^2$ (A manifold I'll call $M$) is diffeomorphic to the Möbuis strip.
One way to see this is using the projection $M\to\mathbb{RP}^1\cong S^1$ given by mapping each line to the unique parallel line which passes through the origin. Each fiber $\pi^{-1}(l)$ of this projection can be identified with the orthogonal compliment of $l$., and the resulting bundle has typical fiber $\mathbb{R}$ and is nontrivial. (If you're familiar, this construction is equivalent to the tautological line bundle of $\mathbb{RP}^1$.)
This set of lines admits a transitive $E(2)$ action, so we have $M\cong E(2)/G$, where $G$ is the stabilizer of the $x$-axis.