Top Lyapunov Exponent of a cocycle is invariant under the map.

Question

Given an map invertible measurable $T$ on a space $(X,\mu)$, take a linear cocycle $\mathcal{A}(x,n)$.

The top Lyapunov exponent at $x$ is defined as $$\chi(x)=\limsup_{n \to \infty}\frac{1}{n}\log{\|\mathcal{A}(x,n)\|}$$

To my understanding $\chi(x)=\chi(T(x))$. I.E the Lyapunov exponent is invariant under the map. Is this obvious? I saw this without proof so I assume this is elementary, but I honestly do not see it.

Thanks!