$x=f(f(x-1))\Rightarrow f(x)=f(f(f(x-1)))=f(f(x-1))+1=(x-1)+1+1=x+1\Leftrightarrow f(x)=x+1$. $x+1=f(f(x))=f(x+1)=x+2\Leftrightarrow 1=2$ So there are no solutions.

$f(f(f(x))=f(x)+1=f(x+1)$ so by induction we get $f(x+n)=f(x)+n$. Setting $x=0$ yields $f(n)=n+f(0)$, so $f$ is linear, but no linear solutions exists.