Assume that $f(x) = (p_1(x),p_2(x),p_2(x))$ is a homogeneous polynomial inducing a diffeomorphic mapping of $\mathbb{R}^3\setminus\{0\}$ onto itself. Homegeneous means $f(tx)=t^n f(x)$, for $t>0$ and some $n\in \mathbb{N}$. Why $n$ should be 1?

Assume that $f(x) = (p_1(x),p_2(x),p_3(x))$ is a homogeneous polynomial inducing a diffeomorphic mapping of $\mathbb{R}^3\setminus\{0\}$ onto itself. Homegeneous means $f(tx)=t^n f(x)$, for $t>0$ and some $n\in \mathbb{N}$. Why $n$ should be 1?

Assume that $f(x) = (p_1(x),p_2(x),p_2(x))$ is a homogenic polynomial diffeomorphic mapping of $R^3\setminus\{0\}$ onto $R^3\setminus\{0\}$ i.e.  $f(tx)=t^n f(x)$, for $t>0$ and some $n\in N$. Why $n$ should be 1?

Assume that $f(x) = (p_1(x),p_2(x),p_2(x))$ is a homogeneous polynomial inducing a diffeomorphic mapping of $\mathbb{R}^3\setminus\{0\}$ onto itself. Homegeneous means $f(tx)=t^n f(x)$, for $t>0$ and some $n\in \mathbb{N}$. Why $n$ should be 1?