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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 $x+y+z=2-1-1=0$).

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    Any $f(x,y,z) = (f_1,f_2,f_3)$ will work, where $f_1,f_2,$ and $f_3$ are linear in $x,y,z$ and they sum to zero. – 2017-01-15

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Besides linearity and the condition $x+y+z=0$, there is one more requirement you need to be sure is satisfied: you need to know that $f$ can achieve every value $(x,y,z)$ such that $x+y+z=0$. For instance, if you define $f(x,y,z)=(0,y,-y)$, then $f$ is linear and $0+y+(-y)=0$, but the image of $f$ is not $\{(x,y,z):x+y+z=0\}$; rather, the image is only a proper subset of this set, since the image does not contain any $(x,y,z)$ such that $x\neq 0$.

So you need to check that given numbers $a,b,c$ such that $a+b+c=0$, then it is always possible to find $x$, $y$, and $z$ such that $f(x,y,z)=(a,b,c)$. For instance, in your first example $f(x,y,z)=(-y-z,y,z)$, this is always possible because you can take $y=b$, $z=c$, and $x$ to be anything.

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This can be done systematically if you work with the matrix $M$ of $f$. We know that $W=\operatorname{im}f$ is spanned by the columns of $M$. $W$ is two-dimensional, so choose for the first two columns any basis of $W$ and for the third any linear combination of the first two. These choices correspond to “playing” with the variables, as you put it.