If $E[X\mid Y]=0$ a.s., does it follow that $E[X\mid f(Y)]=0$ a.s.? If $E[X\mid Y]=0$ a.s., does it follow that $E[X\mid f(Y)]=0$ a.s., for any measurable function $f$?
Let $\mathcal{F}$ be the $\sigma$-algebra generated by $Y$ and $\mathcal{G}$ the $\sigma$-algebra generated by $f(Y)$.  By definition, if $A\in\mathcal{G}$, then there is a Borel set such that $A=(f\circ Y)^{-1}(B)$. But $(f\circ Y)^{-1}=Y^{-1}\circ f^{-1}$ and so 
$$
A=(f\circ Y)^{-1}(B)=Y^{-1}(f^{-1}(B)).
$$ 
Because of $f$'s measurability, $f^{-1}(B)$ is also Borel-measurable. We infer that $A\in\mathcal{F}$ and, more importantly, $\mathcal{G}\subset\mathcal{F}$. Now the claim follows:
$$
E[X|f(Y)]=E[X|\mathcal{G}]=E[E(X|\mathcal{F})|\mathcal{G}]=E[E(X|Y)|\mathcal{G}]=E[0|\mathcal{G}]=0.
$$
The second equality above makes use of iterated conditioning and $\mathcal{G}\subset\mathcal{F}$.
If you take $f(x)=x$ then $E[X|f(Y)]=E[X|Y]=0$. 

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