Consider the infinite cartesian product $\Omega=\{0,1\}^{\mathbb{N}}$
as a measurable space endowed with the $\sigma$-algebra $\mathscr{F}$ generated by the cylinder sets and $\sigma:\Omega\to\Omega$ the left shift map. Denote by $\sigma^n(\mathscr{F})$ the $\sigma$-algebra generated by the family of r.v. $\{X\_i(\omega)=\omega\_i: i\geq n\}$. Suppose that $\mu$ is a Borel probability measure such $\mu(E)\in \{0,1\}$ for any $E\in \cap\_{i\in\mathbb{N}}\sigma^i(\mathscr{F})$.
I would like to know if
$\mu(f|\cap\_{i\in\mathbb{N}}\sigma^i(\mathscr{F}))(x)=\mu(f|\cap\_{i\in\mathbb{N}}\sigma^i(\mathscr{F}))(\sigma x), \mu$ a.s.
for any continuous function $f$.


In such generality I suspect that the answer is no, but I was not able to find a counter example. Of course, the answer is positive if $\mu(E)=\mu(\sigma^{-1}(E))$ for all $E\in \cap\_{i\in\mathbb{N}}\sigma^i(\mathscr{F})$ (which is the case of $\sigma$-invariant measures). I spent some time trying to prove that the triviality hypothesis of $\mu$ implies the shift invariance (for events on the tail $\sigma$-algebra). This would be true if either $E=\sigma^{-1}(E)$ or both set have zero $\mu$ measure. For all the tail events I know this is true, but this does not sound reasonable statement to me in such generality. 



Shift Invariance of Backward Martingales for tail trivial probability measures
As discussed in comments, the main question is not well-posed as it stands, since it depends on a choice of representative (mod $\mu$-a.e. equality) for the conditional expectation.


But as to your subsidiary question "Must $\mu$ be shift-invariant on tail events?", the answer is no.


Let $\nu$ be the probability measure on $\{0,1\}^2$ which assigns probability $1/2$ each to the outcomes $(0,0)$ and $(1,1)$. Let $\mu = \prod\_1^\infty \nu$ be the product of infinitely many $\nu$. So under $\mu$, the random variables $X\_1, X\_3, X\_5, \dots$ are iid coin flips, and a.s. we have $X\_1 = X\_2$, $X\_3 = X\_4$, etc. (You can think of the "experiment" here as "flip infinitely many coins, but write down each flip twice".) Then the tail $\sigma$-algebra is $\mu$-almost trivial, by the Kolmogorov 0-1 law.


Let $E\_i$ be the event $\{X\_{2i-1} = X\_{2i}\}$ and $E = \{E\_i \text{ a.a.}\}= \bigcup\_{n=1}^\infty \bigcap\_{i=n}^\infty E\_i$ be the event that $X\_{2i-1} = X\_{2i}$ for all but finitely many $i$. Clearly $E$ is a tail event and $\mu(E) = 1$. But $\sigma^{-1}(E)$ is the event $F = \{F\_i \text{ a.a.}\}$, where $F\_i = \{X\_{2i} = X\_{2i+1}\}$. The $F\_i$ are independent with $\mu(F\_i) = \frac{1}{2}$, so $\mu(F) = 0$.


To highlight the issue in your question, if we write $\mathcal{T}$ for the tail $\sigma$-algebra and take $f$ to be the constant function $1$, it is correct to say $\mu(f \mid \mathcal{T}) = 1\_E$ a.s. The statement "$1\_E(x) = 1\_E(\sigma x)$ a.s." is false. But it is also correct to say $\mu(f \mid \mathcal{T}) = 1$ a.s. The statement "$1(x) = 1(\sigma x)$ a.s." is true. The definition of conditional expectation says nothing about whether $1\_E$ or $1$ is a "preferred" representative.


