Endomorphism with all eigenvectors

For every non zero x there is $\lambda_{x}$ (unique) so that$f(x)=\lambda_{x}x$. Let’s show that this scalar is independant of x.
Take x and y.
If they are proportionate the result is obvious.
Assume they are not proportionate : $f(x)+f(y)=f(x+y)=\lambda_{x+y}(x+y)$
Hence, $\lambda_{y}y=(\lambda_{x+y}-\lambda_{x})x+\lambda_{x+y}y$
(difference of the two preceding relations).

Finally, we conclude using the liberty of the (x,y) family.

So the solutions are the $\lambda Id$, with $\lambda$ scalars.