No there isn't. Note that $g(x)$ is injective since $g(a)=g(b) \rightarrow f(g(a))=f(g(b)) \rightarrow a^3=b^3 \rightarrow a=b$. Notice that $g(f(g(x)))=g(x^3)=g(x)^2$. This means that $g(-1)=g(-1)^2$, $g(0)=g(0)^2$, and $g(1)=g(1)^2$ by plugging $x = \{-1,0,1\}$. This means that each of $g(-1), g(0), g(1)$ are one of $\{0,1\}$, which means that 2 of them share a value, which is impossible by injectivity, contradiction.