Give an example where $\varphi$ is ergodic but $\varphi^2$ is not

Question

I am trying to show that $\varphi^2 = \varphi \circ \varphi$ is ergodic implies that $\varphi$ is ergodic, but not conversely.

My attempt for ergodicity of $\varphi^2 \Rightarrow $ erodicity of $\varphi$ is $$ P(\varphi^{-2}A) = P(\varphi^{-1}(\varphi^{-1}A))=P(\varphi^{-1} A)=\{0,1\} $$

But for the converse part, I am not sure how to construct a counterexample.