TITLE: Non-linear function on $\mathbb{R}^2$ preserving the origin and maps lines onto lines?
QUESTION [6 upvotes]: Is there an $f:\mathbb{R}^2 \to \mathbb{R}^2$ such that:

$(0,0)\mapsto (0,0)$; and
for any $a,b,c$ with $a^2 + b^2 >0$, the set $A=\{(x,y):ax+by=c\}$ is mapped onto $f(A)=\{(x,y):a'x+b'y=c'\}$ for some $a',b',c'$ with $a'^2 + b'^2 >0$; and
$f$ non-linear?

REPLY [4 votes]: Without surjectivity or some equivalent assumption, such an $f$ can exist.  Take $f(x,y) = (x^3+y,0)$.  If you assume that in addition to some line the image of $f$ contains a point off that line, then $f$ must be linear and invertible.  I posed and proved this for myself at some point before finding a reference and proof on (according to my notes) p.107 of Lyndon's Groups and Geometry.  If you cannot locate this book I can send you a scanned copy of my notes on the proof.