IID does not imply same conditional expectation in general

Question

What is a valid condition for which iid random variables $X,X'$ on $(\Omega,\mathcal{F},\mathbb{P})$ have the same conditional expectation (w.r.t., say, to some non-trivial sub-$\sigma$-algebra $\mathcal{G}$ of $\mathcal{F}$)? Clearly if both are independent of $\mathcal{G}$ this is true. I think this is the most general condition, because if one of them (X) is not independent of $\mathcal{G}$, and, say, the other one (X') is, we can come up with examples where for some $\omega$, $\mathbb{E}_{\mathcal{G}}X(\omega)\neq \mathbb{E}X\equiv\mathbb{E}_{\mathcal{G}}X'(\omega)$, as long as we can construct iid random variables such that one is independent of $\mathcal{G}$ and the other one is not. I think this can be done as follows: consider a random variable $Y$ representing a coin-toss ($1$ is head $0$ tail) of a green coin. Then take a red and a blue coin. Glue the red coin to the green (assume the gluing to have no mechanical effect on the toss outcome, i.e. introducing no bias). Then the blue and the red coin are iid, but while the blue coin toss (X') is independent of the green one (Y), the red coin toss (X) is not (it is in fact fixed by the outcomes of $Y$). So $\mathbb{E}(X|Y)(\omega)\neq \mathbb{E}(X'|Y)=1/2$ since, for example, if $\omega\in\{Y=1\}$, $\mathbb{E}(X|Y)(\omega)= \mathbb{E}(X|Y=1)=1\mathbb{P}(X=1,Y=1)/\mathbb{P}(Y=1)+0=1$ because $\mathbb{P}(X=1,Y=1)=\mathbb{P}(Y=1)=1/2$. Is this example correct?