I have to determine an endomorphism $f$ of $\mathbb{R}^3$ s. t. $\mathrm{Im}(f) = \{ x+y+z=0 \}$.

The easiest solution is, since $x=-y-z$, $f(x,y,z)=(-y-z,y,z)$. 

Anyway, if I consider $z=-x-y$, I get $f(x,y,-x-y)$ or also, if I invert the signs, $f(x,y,z)=(-x,-y,x+y)$ or also $f(x,y,z)=(x+y,-x,-y)$. My question is thus if it is acceptable to "play" with the variables as long as the condition (along the linearity ones) is satisfied (meaning, e.g., $f(1,1,1)=(2,-1,-1)$ which satisfies the condition $x+y+z=2-1-1=0$).