Let $\pi : \mathbb{R}^{n+1} \setminus \{0\} \to \mathbb{RP}^n$ be the projection. Let $V$ be a (smooth) vector field on $\mathbb{R}^{n+1} \setminus \{0\}$. Show that $V$ is $\pi$-related to a vector field on $\mathbb{RP}^n$ if and only if $V_{tx} = tV_x$ for all $x \in \mathbb{R}^{n+1} \setminus \{0\}$ and $t \in \mathbb{R} \setminus \{0\}$.
My first thought is that this has something to to with the fact that $tx$ and $x$ are mapped by $\pi$ to the same element. We assume that $V$ is $\pi$-related to a vector field $W$ on $\mathbb{RP}^n$, i.e.
$T_\pi\circ V=W\circ\pi$,
where $T_\pi$ is the tangent map. Hence we have that
$T_\pi\circ V_{tx}=W\circ\pi(tx)=W\circ\pi(x)=T_\pi\circ V_x$,
i.e.
$V_{tx}(\cdot\circ\pi)=V_x(\cdot\circ\pi)$,
where $V_{tx}$ and $V_x$ are derivations on $\mathbb{R}^{n+1}$.
Does this get me any closer to the goal? What can I do from here?
Please tell me if I should add more information. Any hints will be greatly appreciated.
$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Proj}{\mathbf{P}}$It looks to me that $V(tx) = tV(x)$ is sufficient but not necessary: If $V$ is $\pi$-related to a vector field on $\Reals\Proj^{n}$, and if $V'$ is an arbitrary radial field, then $V + V'$ is $\pi$-related to a vector field on $\Reals\Proj^{n}$ since $\pi_{*}V' = 0$.
Anyway, you're thinking along productive lines (no pun), but need a bit more than the fact that $x$ and $tx$ map to the same point under $\pi$. It probably helps to write the multiplicative action of the group of non-zero reals on $X = \Reals^{n+1} \setminus \{0\}$ as mappings: $$ f_{t}(x) = tx. $$ Then $\pi = \pi \circ f_{t}$ for all $t \neq 0$, and $(f_{t})_{*}(x, v) = (tx, tv)$ for all $(x, v)$ in $TX$.
The claim is, a vector field $V$ on $X$ is $\pi$-related to a vector field on $\Reals\Proj^{n}$ if and only if $\pi_{*}V = \pi_{*}(f_{t})_{*}V$, if and only if $\pi_{*}V(tx) = t\pi_{*}V(x)$.