A function is $T$-invariant if $f(T(x))=f(x)$ for all $x\in X$. In text book: Introduction to Dynamical system by Brin, it defines essentially $T$-invariant: if $f(T(x))=f(x)$ almost every for $x\in X$, i.e. $\mu(\{x\in X: f(T(x))\neq f(x)\}=0$.
If $f$ is $T$-invariant, then $f(T^n(x))=f(T^{n-1}(x))=\dots=f(x)$. I think this property still holds when $f$ is essentially $T$-invariant.
Here is my proof: Note that \begin{align*} \{x\in X:f(T(x))\neq f(x)\}\supset& \{x\in T(X): f(T(x))\neq f(x)\}\\ =& \{x=Ty: f(T^2(y))\neq f(T(y)), y\in X\}. \end{align*} We have \begin{align*} &\mu(\{y\in X: f(T^2(y))\neq f(T(y)\}\\ =&\mu(\{x=Ty: f(T^2(y))\neq f(T(y)), y\in X\})\\ \leq & \mu(\{x\in X:f(T(x))\neq f(x)\})=0. \end{align*} This means that $fT^2=fT$ a.e. If $\mathcal{A}$ is complete then the a.e. property is transitive, so that $fT^2=fT=f$ a.e. Then by induction, one can show that $fT^n=fT^{n-1}=\dots=f$ a.e.
Is the proof right? And does the property still hold?
If $f\circ T=f$ a.e. and $f\circ T=f\circ T^{2}$ a.e. then $f=f\circ T^{2}$ a.e. This does not require completeness of $\mathcal A$: $\{x: (f\circ T)(x)\neq f(x)\}$ and $\{x: (f\circ T^{2})(x)\neq (f\circ T)(x)\}$ are null sets and so their union is a null set.
Also, note that $0=\mu \{x: f(T(x)) \neq f(x)\}=\mu (T^{-1}( \{x: f(T(x)) \neq f(x)\})=\mu \{y: f(T^{2}y) \neq f(Ty)\}$.