Consider the Finsler Minkowski space $(R^n,F)$ and the Euclidean space $(R^n,||.||)$. Consider the Finsler Minkowski indicatrix of radius $r$, that is $$\Sigma(r)=\{x\in R^n\ :\ F(x)=r\}$$ FYI. The indicatrix of a Monkowski Finsler metric is (topologically) a spherical fiber bundle over $R^n$. Furthermore consider the Euclidean sphere of radius $r$, that is $$S^n(r)=\{x\in R^n\ :\ ||x||=r\}.$$ Now if there exists any diffeomorphism (an injective and surjective map whose inverse is also injective and surjective) $$f: (R^n,F)\to (R^n,||.||)$$ that takes the indicatrix to the sphere?
I mean I am asking about the indicatrix of a Monkowski Finsler metric which is (topologically) a spherical fiber bundle over $R^n$ is diffeomorphic to the sphere in $R^n$?
The only things that I found is the diffeomorphism between the punctured spaces as $$f:(R^n\backslash\{0\},F)\to (R^n\backslash\{0\},||.||)$$ s.t. $f(x)= \frac{||x||}{F(x)} x$ and its inverse $h(x)= \frac{F(x)}{||x||}x$.
I am going to assume that $F(x)$ is infinitely smooth (or smooth enough) on $\mathbb{R}^n \setminus \{0\}$. It is continuous everywhere because it is a norm. Per your notations, $\Sigma(r) = \{ x \in \mathbb{R}^n \, : \, F(x) = r\}$ and $S(r) = \{ x \in \mathbb{R}^n \, : \, \|x\| = r \}$. Fix $\Sigma(1)$. Then there exists a large enough $R>0$ such that $\Sigma(1) \, \subset \, D_E(R) = \{ x \in \mathbb{R}^n \, : \, \|x\| < R \}$. Consider an infinitely smooth (rotationally symmetric) bump function $$\lambda \, : \, \mathbb{R}^n \, \to \,\,\mathbb{R}$$ such that \begin{align} &\lambda(x) \equiv 0 \,\,\,\text{ for all } \, x \in \mathbb{R}^n \setminus D_E(2R)\\ 0 \leq &\lambda(x) \leq 1 \,\,\,\text{ for all } \, x \in D_E(2R)\setminus D_E(R)\\ &\lambda(x) \equiv 1 \,\,\,\text{ for all } \, x \in D_E(R)\\ \end{align} Rotationally symmetric means it has the form $\lambda(x) = \kappa\big(\|x\|\big)$ for a smooth function $\kappa(s), \,\, s \in [0,\infty)$ (actually, one can be very specific about the way $\kappa$ looks). Define another similar bump function by setting $$\lambda_{\epsilon}(x) = \lambda\left(\frac{R}{\epsilon}x\right) = \kappa\left(\frac{R}{\epsilon}\|x\|\right) $$ which has the properties \begin{align} &\lambda_{\epsilon}(x) \equiv 0 \,\,\,\text{ for all } \, x \in \mathbb{R}^n \setminus D_E(2\epsilon)\\ 0 \leq &\lambda_{\epsilon}(x) \leq 1 \,\,\,\text{ for all } \, x \in D_E(2\epsilon)\setminus D_E(\epsilon)\\ &\lambda_{\epsilon}(x) \equiv 1 \,\,\,\text{ for all } \, x \in D_E(\epsilon)\\ \end{align} Take $\epsilon > 0$ very small, so small that $D_E(3\epsilon)$ is entirely contained in the interior of $\Sigma(1)$.
Consider the expression $H(x) = \big(1-\lambda(x)\big)\,\|x\| + \lambda(x) \, F(x)$ which by construction is an infinitely smooth function for all points outside the disk $D_E(\epsilon)$. Define the infinitely smooth vector field on $\mathbb{R}^n \setminus D_E(\epsilon)$ $$Y(x) = - \, \frac{\nabla H(x)}{\|\nabla H(x)\|^2}$$ where $\nabla$ is the ordinary gradient in $\mathbb{R}^n$ (i.e. roughly speaking it is the one defined via $\| \cdot\|\,$ ). One can verify that, due to the way $H$ is constructed, $\nabla \, H(x) \neq 0$ for all $x \in \mathbb{R}^n \setminus D_E(\epsilon)$ (recall that $\lambda$ was chosen to be rotationally symmetric and the two norms are convex functions).
Furthermore, define the infinitely smooth vector field $$X(x) = \big(1-\lambda_{\epsilon}(x)\big)\, Y(x)$$ for all $x \in \mathbb{R}^n\setminus D_E(\epsilon)$ and $X(x) = 0$ for $x \in D_E(\epsilon)$. Then $X(x)$ is an infinitely smooth vector field on the whole space $\mathbb{R}^n$.
Let $\phi^t(x)$ be the flow of the vector field $X$, i.e. $$\frac{d}{dt} \, \phi^t(x) = X\big(\phi^t(x)\big)$$ Moreover, due to the fact that the vector filed $X$ is bounded everywhere on $\mathbb{R}^n$, i.e. there exists a large enough cpnstant $M>0$ such that $$\|X(x)\| = \big|\big(1-\lambda_{\epsilon}(x)\big)\big| \, \|Y(x)\| \leq M$$ the flow extends for all $t \in \mathbb{R}$. By construction, given $x \in \mathbb{R}^n \setminus D_E(3\epsilon)$ the following relation holds $$H\big(\phi^t(x)\big) = H(x) - t$$ for $t \in (-\infty, H(x) - 2\epsilon)$. Indeed, for any point $x \in \mathbb{R}^n \setminus D_E(3\epsilon)$ we have that $X(x) = Y(x)$ so whenever $t \in (-\infty, H(x) - 2\epsilon)$ \begin{align} H\big(\phi^t(x)\big) - H(x) &= \int_{0}^t \, \frac{d}{d\tau} \, H\big(\phi^{\tau}(x)\big) \, d\tau = \int_{0}^t \, \Big( \, \nabla H\big(\phi^{\tau}(x)\big) \, \cdot \, \frac{d}{d\tau} \, \phi^{\tau}(x) \, \Big) \, d\tau\\ &= \int_{0}^t \, \Big( \, \nabla H\big(\phi^{\tau}(x)\big) \, \cdot \, X\big(\phi^{\tau}(x)\big) \, \Big) \, d\tau\\ &= \int_{0}^t \, \Big( \, \nabla H\big(\phi^{\tau}(x)\big) \, \cdot \, Y\big(\phi^{\tau}(x)\big) \, \Big) \, d\tau\\ & = - \, \int_{0}^t \,\frac{ \Big( \, \nabla H\big(\phi^{\tau}(x)\big) \, \cdot \, \nabla H\big(\phi^{\tau}(x)\big) \, \Big)}{\|\nabla H\big(\phi^{\tau}(x)\big) \|^2} \, d\tau\\ & = - \, \int_{0}^t \,\frac{\|\nabla H\big(\phi^{\tau}(x)\big) \|^2}{\| \nabla H\big(\phi^{\tau}(x)\big) \|^2} \, d\tau = - \, \int_{0}^t \, d\tau\\ & = - t \end{align} The phase flow $\phi^t(x)$ is a diffeomorphism for any fixed $t \in \mathbb{R}$ and can move any level hyper-surface $H(x) = r_1$ to any level hyper-surface $H(x) = r_2$ as long as $r_1 > 2\epsilon$ and $r_1 > 2\epsilon$.
Then, by construction $$\{x \in \mathbb{R}^n \, : \, H(x) = 3R\,\} = S(3R) \,\,\,\,\ \text{ and } \,\,\,\, \{x \in \mathbb{R}^n \, : \, H(x) = 1\,\} = \Sigma(1)$$ Set $t_0 = 3R - 1$. Then, again by construction, $\phi^{t_0}\big(S(3R)\big) = \Sigma(1)$. Then the map $$\phi = \phi^{t_0}|_{S(3R)} \, : \, S(3R) \, \to \, \Sigma(1)$$ is a smooth diffeomorphism between the two submanifolds $S(3R)$ and $\Sigma(1)$ and in fact $$\phi^t(x) \, : \, t \in [0, 3R-1]$$ is a smooth isotopy of $\mathbb{R}^n$ which isotopes $S(3R)$ to $\Sigma(1)$. To get an ambient isotopy from $S(1)$ to $\Sigma(1)$, simply precompose $\phi^t(x)$ with a linear homothetic isotopy that stretches homogeneously $\mathbb{R}^n$, taking $S(1)$ to $S(3R)$.