Let us only consider the 2 dimensional case where the two frames of reference coincide at the origin. For each $v\in (-c,c)$, let $f_v:\mathbb{R}^2\to\mathbb{R}^2$ be a smooth function satisfying the following conditions (with the first coordinate representing time and the second representing space).
$(1)\,\,f_v(0,0)=(0,0)$,
$(2)\,\,\forall t\in\mathbb{R},\exists t'\in\mathbb{R}$ such that $f_v(t,vt)=(t',0)$,
$(3)\,\,\forall t\in\mathbb{R},\exists t'\in\mathbb{R}$ such that $f_v(t,0)=(t',-vt')$,
$(4)\,\,\forall t\in\mathbb{R},\forall x\in\mathbb{R},f_v(t,x)=((\pi_1\circ f_{-v})(t,-x),-(\pi_2\circ f_{-v})(t,-x))$, and
$(5)\,\,\forall t\in\mathbb{R},\exists t'\in\mathbb{R}$ such that $f_v(t,ct)=(t',ct')$.
Are there other hypotheses needed in order to show that $f_v$ is linear? I can show that for all $v\in (-c,c)$ if $f_v$ is a polynomial, then $f_v$ is linear; however, I cannot show that if $f_v$ is a limit of polynomial functions, then $f_v$ is linear.
Most of my attempts trying to figure this out have started with $"$let $$f_v(t,x)=(\sum_{i=0,\, j=0}^{\infty}a_{ij}t^i x^j, \sum_{i=0,\, j=0}^{\infty}b_{ij}t^i x^j)."$$ I have then tried to show that some relationship between the coefficients implies that after a certain point, they are all $0$ (to no avail).