Any diffeomorphism between the Minkowski indicatrix and the Euclidean sphere? 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 $R&gt;0$
such that $\Sigma(1) \, \subset \, D_E(R) = \{ x \in \mathbb{R}^n
\, : \, \|x\| &lt; R \}$. Consider an infinitely smooth
(rotationally symmetric) bump function
$$\lambda \, : \, \mathbb{R}^n \, \to \,\,\mathbb{R}$$ such that
\begin{align}
&amp;\lambda(x) \equiv 0 \,\,\,\text{ for all } \, x \in \mathbb{R}^n \setminus D_E(2R)\\
0 \leq &amp;\lambda(x) \leq 1 \,\,\,\text{ for all } \, x \in D_E(2R)\setminus D_E(R)\\
&amp;\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}
&amp;\lambda_{\epsilon}(x) \equiv 0 \,\,\,\text{ for all } \, x \in \mathbb{R}^n \setminus D_E(2\epsilon)\\
0 \leq &amp;\lambda_{\epsilon}(x) \leq 1 \,\,\,\text{ for all } \, x \in D_E(2\epsilon)\setminus D_E(\epsilon)\\
&amp;\lambda_{\epsilon}(x) \equiv 1 \,\,\,\text{ for all } \, x \in D_E(\epsilon)\\
\end{align}
Take $\epsilon &gt; 0$ 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 $M&gt;0$ 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) &amp;= \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\\
&amp;= \int_{0}^t \, \Big( \, \nabla H\big(\phi^{\tau}(x)\big) \, \cdot \, X\big(\phi^{\tau}(x)\big) \, \Big) \, d\tau\\
&amp;= \int_{0}^t \, \Big( \, \nabla H\big(\phi^{\tau}(x)\big)
\, \cdot \, Y\big(\phi^{\tau}(x)\big) \, \Big) \, d\tau\\
&amp; = - \, \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\\
&amp; = - \, \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\\
&amp; = - 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 $r_1 &gt; 2\epsilon$ and $r_1 &gt;
2\epsilon$.

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 $S(3R)$.
Yes, the indicatrix of a Minkowski space is diffemorphic to a sphere. 

Let $I$ and $S$ be the indicatrix of a Minkowski (V,F) and the sphere of a Euclidian space, resp. The aplication $f:S\to I$ give by $$v\mapsto \frac{v}{F(v)}$$
is a diffomorphism. For more details vide the Proposition 2.6 on [Javaloyes and Sànches, On the definiton and examples of Finsler metrics].