Are all collinearity preserving bijections from $\newcommand{\RR}{\mathbb R}\RR^3$ to $\mathbb R^3$ of the form $Ax+b$ for $A$ in $M_{3\times3}(\RR)$ and $b \in \mathbb R^3$?
In particular, let bijective $f:\mathbb R^3\to\mathbb R^3$ such that if $a,b,c$ in $\mathbb R^3$ are collinear then $f(a),f(b),f(c)$ are collinear. Prove or disprove that there exists $A$ in $M_{3\times3}(\mathbb R)$ and $b$ in $\mathbb R^3$ such that $f(x)=Ax+b$.
My work so far:
Because $f$ is bijective, $f^{-1}$ exists, thus we can find
$a = f^{-1}((0,0,0))$, $b = f^{-1}((1,0,0))$, $c = f^{-1}((0,1,0))$, $d = f^{-1}((0,0,1))$
and construct the unique linear transform T such that
$a = T((0,0,0))$, $b = T((1,0,0))$, $c = T((0,1,0))$, $d = T((0,0,1))$
notice $g$ fixes $(0,0,0), (1,0,0), (0,1,0), (0,0,1)$
Then let $g(x) = f(T(x))$, if we can show that $g$ is of the form $Ax+b$ then because $f(x) = g(T^{-1}(x))$ and $g$ and $T$ are both of the form $Ax+b$, then we will have show $f$ is of the form $Ax+b$, thus it is sufficient to consider a collinearity preserving bijection that fixes the origin and 3 standard basis vectors.
I'm not sure where to go from here, any help is appreciated, thanks.