Projective transformation

Question

I need to find the function $f$ that satisfies the following:

$f((1:1:0))=(0:1:1)$

$f((0:1:1))=(1:0:1)$

$f((1:0:1))=(1:1:0)$

If I let:

$x=(1:1:0)$

$y=(0:1:1)$

$z=(1:0:1)$,

then I get $f(x)=y$, $f(y)=z$, $f(z)=s$.

So it has somewhat of a cyclic property, where $f(f(f(x)))=x$.

But I do not have any idea to find the function. Could we use a matrix, where $A^3x=x$, and invert it?

Answer 1

$\pmatrix{0 &0 &1 \\ 1&0&0\\0&1&0} $

You can solve this just by looking at it for one minute. See what happens to common elements... eg: $f((1,\textbf{1},0))=(0,1,\textbf{1}) $

$f((0,\textbf{1},1))=(1,0,\textbf{1}) $

and so on...