When is $\sigma(E[X|\mathcal F]) \subset \sigma(E[Y|\mathcal F])$? Let $X,Y$ be integrable random variables on the same probability space and $\mathcal F$ a sub-sigma algebra.
Under what conditions on $X,Y$ do we have $\sigma(E[X|\mathcal F]) \subset \sigma(E[Y|\mathcal F])$ ? Is it $\sigma(X) \subset \sigma(Y)$ ?
 A: I don't know a natural condition for this to be true, but $\sigma(X) \subset \sigma(Y)$ isn't sufficient.
Let $\Omega = \{a,b,c,d\}$ be a sample space with 4 points, all having probability 1/4, and $\sigma$-algebra $2^\Omega$.
Define random variables $X,Y$ by
$$X(a) = 0, \quad X(b) = 2, \quad X(c) = 4, \quad X(d) = 6$$
$$Y(a) = -1, \quad Y(b) = 1, \quad Y(c) = -2, \quad Y(d) = 2$$
Since $X,Y$ are both injective, we have $\sigma(X) = \sigma(Y) = 2^\Omega$; in particular $\sigma(X) \subseteq \sigma(Y)$. Now let $\mathcal{F} = \{\{a,b\}, \{c,d\}, \emptyset, \Omega\}$.  You can directly compute the conditional expectations and find that
$$E[X \mid \mathcal{F}] = 1_{\{a,b\}} + 5_{\{c,d\}}, \quad E[Y \mid \mathcal{F}] = 0.$$
Thus $\sigma(E[Y \mid \mathcal{F}])$ is trivial and $\sigma(E[X \mid \mathcal{F}])$ is not, so we do not have $\sigma(E[X \mid \mathcal{F}]) \subset \sigma(E[Y \mid \mathcal{F}])$.
The condition isn't necessary either.  Let $X,Y$ be any two random variables on any probability space for which $\sigma(X) \not\subseteq \sigma(Y)$, and let $\mathcal{F} = \{\Omega, \emptyset\}$ be trivial.  Then $E[X \mid \mathcal{F}]$ and $E[X \mid \mathcal{G}]$ are both constant, so their generated $\sigma$-fields are trivial and thus equal, so in particular $\sigma(E[X \mid \mathcal{F}]) \subseteq \sigma(E[Y \mid \mathcal{F}])$.
The basic problem is that the $\sigma$-field $\sigma(X)$ doesn't really care about the numerical values taken on by $X$; it only looks at the codomain $\mathbb{R}$ of $X$ as a measurable space, with no relevant algebraic structure.  But $E[X \mid \mathcal{F}]$ depends greatly on the actual numerical values, and uses the vector space structure of $\mathbb{R}$ in its definition.  So you will not learn much about $E[X \mid \mathcal{F}]$ by looking only at $\sigma(X)$, and therefore you will not learn much about $\sigma(E[X \mid \mathcal{F}])$ either.
To say it another way, if you replace $X$ by $g(X)$ for some injective function $g$, then the generated $\sigma$-field doesn't change; but $E[X \mid \mathcal{F}]$ can be completely different from $E[g(X) \mid \mathcal{F}]$, and so their generated $\sigma$-fields can likewise be completely different.
