If $Y$ and $X$ are independent random variables, we have
$$\mathbb{E}(Y\mid X)=\mathbb{E}(Y) \tag{1}$$
$$\mathbb{E}(X\mid Y)=\mathbb{E}(X) \tag{2}$$
How about the converse? Obviously, neither $(1)$ nor $(2)$ alone implies that $Y$ and $X$ are independent.
But suppose $(1)$ and $(2)$ both hold. Does this imply that $Y$ and $X$ are independent? Can anyone supply a counterexample?
Suppose $(X,Y)$ is uniformly distributed on a unit disk with density
$$f_{X,Y}(x,y)=\frac{\mathbf1_{x^2+y^2<1}}{\pi}$$
Then $X$ and $Y$ are identically distributed, with marginal density of $X$ being
$$f_X(x)=\left(\frac{2}{\pi}\sqrt{1-x^2}\right)\mathbf1_{|x|<1}$$
Due to symmetry, $$\mathbb E\left[X\right]=\mathbb E\left[Y\right]=0$$
Conditional density of $Y$ given $X=x$ is
$$f_{Y\mid X}(y\mid x)=\frac{f_{X,Y}(x,y)}{f_X(x)}=\frac{\mathbf1_{x^2+y^2<1,|x|<1}}{2\sqrt{1-x^2}}$$
In other words, $Y$ given $X=x$ is uniform on $\left(-\sqrt{1-x^2},\sqrt{1-x^2}\right)$ for $|x|<1$.
Similarly $X$ given $Y=y$ is uniform on $\left(-\sqrt{1-y^2},\sqrt{1-y^2}\right)$ for $|y|<1$
Again due to symmetry, this implies $$\mathbb E\left[Y\mid X\right]=0 \,,\text{ a.e. }$$ and $$\mathbb E\left[X\mid Y\right]=0 \,,\text{ a.e. }$$
So $\mathbb E\left[Y\mid X\right]=\mathbb E\left[Y\right]$ a.e. and $\mathbb E\left[X\mid Y\right]=\mathbb E\left[X\right]$ a.e. But still $X$ and $Y$ are not independent.