For matrices this is true, but I'm not sure whether it applies to more general transformations. Here $T$ is a finite measure-preserving transformation. From the definition of eigenvalues, if $\lambda$ is an eigenvalue of $T$ then there exists a nonzero a.e. function $f \in L^2$ such that $f(T(x)) = \lambda(f(x))$ a.e.
Then we have that $f(T^2(x)) = f(T(T(x)) = \lambda f(T(x)) = \lambda \cdot \lambda f(x) = \lambda^2 f(x)$. Does this logic hold?
One has to add one detail: let $A = \{ x : f(T(x)) = \lambda f(x) \}$. Then $A$ is of full measure and $f(T^2(x)) = \lambda^2 f(x)$ whenever $x \in A$ and $T(x) \in A$, i.e. $x \in A \cap T^{-1}[A]$. Since $T$ is measure-preserving, this set is also of full measure, so the equality holds almost everywhere.