Let $\cal H$ be the $\sigma$-algebra generated by ${\Bbb E} (X\mid\mathcal G)$.
As you say, since ${\Bbb E} (X\mid\mathcal G)$ is $\mathcal G$-measurable, $\cal H$ is contained in $\mathcal G$. It doesn't have to be contained in $\sigma(X)$. For a simple counterexample, let the probability space be $\Omega=\{1,2,3,4\}$ with all subsets measurable and the uniform probability measure, and let $${\mathcal G}=\{\emptyset,\{1,2\},\{3\},\{4\},\{3,4\},\{1,2,3\},\{1,2,4\},\Omega\},$$ $$X(1)=1, X(2)=X(3)=X(4)=0.$$ Then
$$
{\Bbb E} (X\mid\mathcal G)(1)={\Bbb E} (X\mid\mathcal G)(2)=\frac12,
\ \ \ \
{\Bbb E} (X\mid\mathcal G)(3)={\Bbb E} (X\mid\mathcal G)(4)=0$$
so ${\cal H}=\{\emptyset,\{1,2\},\{3,4\},\Omega\}$ is not contained in $\sigma(X)=\{\emptyset,\{1\},\{2,3,4\},\Omega\}$.
Since $\cal H$ does not have to be contained in $\sigma(X)$, it need not equal ${\cal G}\cap\sigma(X)$. The example above shows that it also does not have to equal $\cal G$.
Addendum: As Byron Schmuland says below, if $X$ is an ${\cal F}/{\cal B}({\Bbb R})$-measurable real-valued function, then $\cal H$ is countably generated as it's the pullback of the countably generated $\sigma$-algebra ${\cal B}(\Bbb R)$ under ${\Bbb E} (X\mid\mathcal G)^ {-1}$. Also, any countably generated $\sigma$-algebra which is contained in $\cal G$ can occur as a value of $\cal H$ for some real-valued $X$. (${\Bbb E} (X\mid\mathcal G)$ is only defined up to equality on a set of measure $1$, so to be precise you should say that any countably generated $\sigma$-algebra which is contained in $\cal G$ can occur as a value of $\sigma({\Bbb E}(X\mid\mathcal G))$ for some real-valued $X$ and some version of ${\Bbb E}(X\mid \mathcal G)$.)
If a probability space is given then the conditional expectation does not depend on the values taken by the random variable in the condition; it depends only on the sets on which this random variable takes these values as constants. You can experience this if you calculate a very simple example when the random variables involved are discrete.
Since only the sets count, one can say that only the indicator functions count.
So, if you have a $(\sigma$-)algebra in the condition then you can replace this
$\ \ (\sigma-)$algebra by any random variable which generates the same $(\sigma$-)algebra. Or the other way around, if you have a random variable in the condition then you can replace it by the $(\sigma$-)algebra generated by that random variable.
To answer your question briefly: If a random variable is in the condition of a conditional expectation then it can be considered as a system of indicator functions of sets on which the random variable is constant.
Best Answer
For any $\eta \in L^{\infty}$, $$E(X\eta) = E(XE(\eta \mid \mathscr{G})) = E(E(X \mid \mathscr{G})E(\eta \mid \mathscr{G})) = E(E(X \mid \mathscr{G})\eta).$$ Taking $\eta$ to be suitable indicator functions, this implies that $X = E(X \mid \mathscr{G})$ a.s.. So $X$ is a.e. equal to a $\mathscr{G}$-measurable function.