I am wondering about the problem of finding for a fixed $\sigma$-algebra $\mathcal F$ and a fixed random variable, the smallest $\sigma$-algebra $\mathcal G$ such that $\mathbb E[X\mid\mathcal F]=\mathbb E[X\mid\mathcal G]$ almost surely, a candidate that came to mind is $\mathcal G=\mathcal F\cap \sigma(X)$. In order to prove it we could show that $\mathbb E[X\mid\mathcal G]$ is $\mathcal F$ measurable which is obvious. Then we would have to show that for any $\mathcal F$-measurable random variable $U$, $\mathbb E[XU]=\mathbb E[\mathbb E[X\mid\mathcal G]U]$.
EDIT: According to GEdgar's answer, $\mathcal G$ is not $\mathcal F\cap \sigma(X)$. Any thought on what $\mathcal G$ should be is most welcome.
Example. Let $$ \Omega = \{0,1\} \times \{0,1\} $$ Let $\mathbb P$ be uniform measure $$ \mathbb P\{(0,0)\} =\mathbb P\{(0,1)\} =\mathbb P\{(1,0)\} =\mathbb P\{(1,1)\} =\frac{1}{4} $$ Let random variable $X$ be $X(s,t) = s-t$, $$ X(0,0) = X(1,1) = 0,\qquad X(0,1) = -1,\qquad X(1,0) = 1. $$ Let $\mathcal F$ have two atoms $$ \{0\}\times\{0,1\},\qquad\{1\}\times\{0,1\} $$ So $\mathcal F$ has four elements: those two atoms, $\Omega$, and $\varnothing$.
Conditional expectation $C := \mathbb E[X\mid \mathcal F]$ is $$ C(0,0) = C(0,1) = \frac{(-1)+0}{2} = \frac{-1}{2},\\ C(1,0) = C(1,1) = \frac{0+1}{2} = \frac{1}{2} $$ Sigma-algebra $\sigma(X)$ has three atoms: $$ \{(0,0), (1,1)\},\quad \{(0,1)\},\quad \{(1,0)\} $$ Sigma-algebra $\mathcal F \cap \sigma(X)$ is trivial: $$ \mathcal F \cap \sigma(X) = \{\Omega,\varnothing\} $$ Thus $\mathbb E[X \mid \mathcal F]$ is not $\mathcal F \cap \sigma(X)$-measurable.