We know that if $f(x+y)=f(x)+f(y)$ and $f$ meets some "reasonable" conditions, then $f$ is linear.
I've been considering the following extension: consider the reals under some unknown group operation $\oplus$ which is isomorphic to the reals under standard addition, i.e. $f(x\oplus y)=f(x)+f(y)$. Under what conditions must we conclude that $f(x)=cx$ where $f$ is the isomorphism?
I think over the rationals we could use the same argument as for Cauchy, but I'm not sure about over the reals.
Update: In another question Joy gives an example where $f:(\mathbb{R},\oplus)\to (\mathbb{R},+)$ with $f$ continuous yet $f(x)\not=cx$. So the answer to this generalization is not the same as the answer to the normal Cauchy.
If you conclude $f(x)=cx$, then $c(x\oplus y)=f(x\oplus y)=f(x)+f(y)=cx+cy=c(x+y)$, that is, $\oplus=+$. Now for an automprhism of $f$ of $(\mathbb{R},+)$, to conclude that $f(x)=cx$ then $f$ must be continuous, or equivalently, $\lim_{x\rightarrow 0}f(x)=0$.