We call a normed vector space $E$ a Newtonian space if $E$ is isometrically isomorphic to $\mathbb R^3$. With $E$ a Newtonian space, the space $E\times\mathbb R$ is called a Newtonian spacetime.
Suppose $E$ is a Newtonian space and $I\subseteq\mathbb R$. A free particle in Newtonian spacetime is a map \begin{align} p:\ I&\longrightarrow E \\ s&\longmapsto p(s)=as+b \end{align} with some fixed $a,b\in E$. The set $$W=\big\{(p(s),s)\,|\,s\in I \big\}\subset E\times\mathbb R $$ is called world line of the particle $p$.
If a diffeomorphism \begin{align} (x,t):\ E\times\mathbb R&\longrightarrow\mathbb R^4 \\ (p,s)&\longmapsto \big(x(p,s),t(p,s)\big) \end{align} satisfies $x(W)$ is a straight line, for every free particle, then we say $x$ is an inertial frame for the Newtonian spacetime.
I wonder that if $x$ is an inertial frame, and we consider its spatial component \begin{align} x:\ E&\longrightarrow\mathbb R^3 \\ p&\longmapsto x(p) = x(p,s) \end{align} then would $x$ be a bijective isometry ?
My attempt is to show that $x$ bring a straight line $p(s) = as+b$ of $E$ into a straight line of $\mathbb R^3$, then $x$ should be an affine map. But I have no idea how to show it. Could anyone give me a little help for this problem ? I really appreciate it.