$\newcommand{\mc}{\mathcal}$ $\newcommand{\E}{\mathbf E}$ $\newcommand{\R}{\mathbf R}$ $\newcommand{\P}{\mathbb P}$ $\newcommand{\wh}{\widehat}$
Definitions
Let $G$ be a connected graph on a finite set of vertices $X$ and let $P$ denote the random walk on $G$. For a vertex $y\in X$ and for $E\subseteq X$, we write $P_y(E)$ to denote the probability of landing in $E$ in the next step given that we are at $y$.
We use $\mc X$ to denote the power set of $X$. For each $n\geq 0$, let $\Omega_n=\prod_{i=0}^n X$ and equip it with the product $\sigma$-algebra (which is same as the power set of $\Omega_n$). Let $x$ be a point in $X$. Write $\P_x^0$ to denote the Dirac measure $\delta_x$ on $\Omega_0=X$.
We define a measure $\P_x^1$ on $\Omega_1$ as follows: $$ \P_x^1(E_0\times E_1) = \int_{E_0} P_y(E_1) \ d\P_x^0(y) $$ for all $E_0, E_1\in \mc X$. Intuitively, $\P_x^1(E_0\times E_1)$ is the probability that the random walk ``starts at $x$ and follows the trajectory $E_0\times E_1$".
More generally, once $\P_x^n$ is constructed, we can construct $\P_x^{n+1}$ by declaring $$ \P_x^{n+1}(E_0\times \cdots \times E_{n+1}) = \int_{E_0\times \cdots \times E_n} P_{\omega_n}(E_{n+1})\ d\P_x^n(\omega) $$ for all $E_0 , \ldots, E_{n+1}\in \mc X$. Again, intuitively, $\P_x^{n+1}(E_0\times \cdots \times E_{n+1})$ is the probability that the random walk follows the trajectory $E_0\times \cdots \times E_{n+1}$ provided it starts at $x$.
Note that the push-forward of $\P_x^{n}$ to $\Omega_{n-1}$ by the projection map $\Omega_n\to \Omega_{n-1}$ which kills the last coordinate is same as $\P_x^{n-1}$. So we have a unique probability measure $\P_x$ on the product space $\Omega=\prod_{n=0}^\infty X$ (equipped with the product $\sigma$-algebra) such that the push-forward of $\P_x$ to $\Omega_n$ under the projection map $\Omega\to \Omega_n$ which kills all coordinates beyond $n$ is $\P_x^n$. See the figure below where all the arrows are projection maps.
Now suppose $\mu$ is a probability distribution on $X$. We define a probability measure $\P_\mu$ on $\Omega$ as $$ \P_\mu = \int_X \P_x\ d\mu(x) $$ The message is that $\P_\mu(E_0\times \cdots \times E_n)$ is the probability that the chain follows the trajectory $E_0\times \cdots \times E_n$ provided it starts according to the distribution $\mu$; we start at $x$ with probability $d\mu(x)$ and once we are at $x$ we follow a given trajectory with probability $\P_x$.
Define a map $T:\Omega\to \Omega$ by setting $$ T(\omega_0, \omega_1, \omega_2, \ldots) = (\omega_1, \omega_2, \omega_3, \ldots) $$ The map $T$ is called the one-sided shift.
It is a simple fact that if $\mu$ is a stationary distribution of the random walk (in fact, since the graph is connected, there is only one stationary distribution), then the measure $\P_\mu$ is $T$-invariant, that is, $T_*\P_\mu=\P_\mu$.
Question
Let $\mc X_n$ be the $\sigma$-algebra on $\Omega$ obtained by pulling back the product $\sigma$-algebra on $\Omega_n$ via the projection map $\pi_n:\Omega\to \Omega_n,\ \omega{\mapsto} (\omega_0 , \ldots, \omega_n)$. Let $\mu$ be the stationary distribution of the random walk. For any bounded measurable function $\psi:\Omega\to \R$, we will write $\E[\psi|\mc X_n]$ to denote the conditional expectation of $\psi$ given $\mc X_n$ with respect to the measure $\P_\mu$.
I want to prove the following
Lemma 1. For any bounded measurable function $\psi$ on $\Omega$ we have $$ \E[\psi\circ T^n|\mc X_n]=\E[\psi|\mc X_0]\circ T^n $$
I am able to see the above only partially even in the case $n=1$.
Let $E_0, E_1$ be Borel sets in $X$. Write $\wh E_0$ to denote $\pi_0^{-1}(E_0)$ and $\wh E_1$ to denote $\pi_0^{-1}(E_1)$. Also write $S$ to denote $\pi_1^{-1}(E_0\times E_1)$. Then we have for any bounded measurable function $\psi:\Omega\to \R$ that \begin{align} \begin{split} \int_{\wh E_1} \E[\psi|\mc X_0]\ d\P_\mu &= \int_{\wh E_1} \psi\ d\P_\mu\\ \Rightarrow \int_{X\times \wh E_1} \E[\psi|\mc X_0]\circ T\ d\P\mu &= \int_{X\times \wh E_1}\psi \circ T\ d\P_\mu \end{split} \end{align} But I am not able to show the following $$ \int_{S} \E[\psi|\mc X_0]\circ T\ d\P_\mu = \int_S \psi\circ T\ d\P_\mu $$ The last equation as such is not enough to prove the lemma since not all members of $\mc X_1$ look like rectangles. But a little bit of extra work from here will actually show the desired.
I don't have a full solution either, but perhaps some of these thoughts might help you to find one...
Let $\mathcal{B}_n$ be the $\sigma$-algebra obtained by pulling back the $\sigma$-algebra on $\Omega$ via the map $\Omega\to X, \omega\mapsto \omega_n$. By definition $B_0=\mathcal{X}_0$, and $\mathcal{B}_n \subset \mathcal{X}_n$.
I claim that $$\mathbf{E}[\psi| \mathcal{X}_0]\circ T^n= \mathbf{E}[\psi \circ T^n| \mathcal{B}_n].$$
(so what you are looking for reduce to prove that $\mathbf{E}[\psi \circ T^n| \mathcal{B}_n]=\mathbf{E}[\psi \circ T^n| \mathcal{X}_n]$, in other words, is $\mathbf{E}[\psi \circ T^n| \mathcal{X}_n]$ measurable wrt $\mathcal{B}_n$ ?)
To prove this, first note that a map $f:X\to \mathbf{C}$ is $\mathcal{B}_n$-measurable iff it depends only on the $n$-th coordinate; so for any such $f$, there exists $F:X\to \mathbf{C}$, $\mathcal{X}_0$-measurable such that $f=F\circ T^n$.
Now, the function $\mathbf{E}[\psi|\mathcal{X}_0]\circ T^n$ is clearly $\mathcal{B}_n$ measurable. To show that it is the conditional expectation of $\psi\circ T^n$ wrt $\mathcal{B}_n$, we need to compute its integral versus an arbitrary $g$ which is $\mathcal{B}_n$-measurable, i.e of the form $g=G\circ T^n$.
$$I:=\int_\Omega g(\mathbf{E}[\psi| \mathcal{X}_0]\circ T^n) d\mathbb{P}_\mu(\omega)=\int_\Omega G \circ T^n \mathbf{E}[\psi| \mathcal{X}_0]\circ T^n d\mathbb{P}_\mu,$$ by $T$-invariance we have $$I=\int G \mathbf{E}[\psi| \mathcal{X}_0] d\mathbb{P}_\mu,$$ Now, since $G$ is $\mathcal{X}_0$-measurable, by definition of conditional expection $$I=\int G \psi d\mathbb{P}_\mu=\int G\circ T^n \psi\circ T^n d\mathbb{P}_\mu, $$ again using $T$-invariance of $\mathbb{P}_\mu$, so
$$I=\int g \psi \circ T^n d\mathbb{P}_\mu,$$ as required.