Let $(E,\mathcal E,\mu)$ denote the Lebesgue measure space on $[0,1)$, $$\tau(x):=2x-\lfloor 2x\rfloor\;\;\;\text{for }x\in E,$$ $$Y_0:=\lfloor 2x\rfloor\;\;\;\text{for }x\in E$$ and $$Y_n:=Y_0\circ\tau^{n-1}\;\;\;\text{for }n\in\mathbb N.$$ We can show that $\mu\circ\tau^{-1}=\mu$ and $(Y_n)_{n\in\mathbb N}$ is an independent and identically distributed process on $(E,\mathcal E,\mu)$. Let $$(\Omega,\mathcal A,\operatorname P):=\left(E^{\mathbb N},\mathcal E^{\otimes\mathbb N},\left(\mu\circ Y_1^{-1}\right)^{\otimes\mathbb N}\right),$$ $\theta$ denote the shift on $\Omega$, i.e. $$\theta:\Omega\to\Omega\;,\;\;\;\omega\mapsto(\omega_{n+1})_{n\in\mathbb N}$$ and $$\varphi(\omega):=\sum_{n\in\mathbb N}\frac{\omega_n}{2^n}\;\;\;\text{for }\omega\in\Omega.$$ We can show that $$\tau^n=\varphi\circ\theta^n\circ Y\;\;\;\text{for all }n\in\mathbb N_0.\tag1$$
Let $$\mathcal I_\tau:=\{B\in\mathcal E:\tau^{-1}(B)=B\}$$ and $$\mathcal I_\theta:=\{A\in\Omega:\theta^{-1}(A)=A\}.$$ We know that $\operatorname P\circ\:\theta=\operatorname P$ and that $\mathcal I_\theta$ is $\operatorname P$-trivial. Are we able to conclude that $\mathcal I_\tau$ is $\mu$-trivial?
Let $B\in\mathcal I_\tau$. Then, \begin{equation}\begin{split}\mu(B)&=\mu(\tau\in B)=\mu(\varphi\circ\theta\circ Y\in B)\\&=\mu\left(Y\in\theta^{-1}\left(\varphi^{-1}(B)\right)\right)\\&=\operatorname P\left[\theta^{-1}\left(\varphi^{-1}(B)\right)\right].\end{split}\tag2\end{equation} So, it seems like we would need that $\varphi^{-1}(B)\in\mathcal I_\theta$ which boils down to showing that $\varphi$ is $(\mathcal I_\theta,\mathcal I_\tau)$-measurable.
Am I missing something? If not, how can we do that?
Well, we know $\varphi$ is $(\mathcal{A},\mathcal{E})$-measurable, so given any $B \in \mathcal{I}_\tau$, we know $\varphi^{-1}(B) \in \mathcal{A}$, so to show $\varphi^{-1}(B) \in \mathcal{I}_\theta$, you just need to show $\theta^{-1}(\varphi^{-1}(B)) = \varphi^{-1}(B)$, and I think you've already done this.