Let $X \overset{D}{=} X_{1}$ be an independent copy of a continuous random variable $X$, where $X$ is on probability space $(\Omega, F, \mathrm{P})$. Let $A$ be another random variable, where $E (A\mid X)$ is known. Further, $X$ and $X_{1}$ depend on $A$. I am trying to show that \begin{equation} E (A\mid X) = E (A\mid X_{1}). \end{equation} If counterexamples exist, please state them.
Consider $G_1 \subseteq G_2 \subseteq F$, where $G_1= [\emptyset, \Omega]$ is the trivial sigma-algebra and $G_2=\sigma(X)$.
My attempt is the use the law of iterated expectation \begin{equation} \mathrm{E}\left[\mathrm{E}\left[X \mid G_2\right] \mid G_1\right]=\mathrm{E}\left[X \mid G_1\right], \end{equation} precisely \begin{equation} E (A\mid X_{1}) = E(E(A\mid X)\mid X_{1}). \end{equation} to show the statement of interest. I believe that $G_2=\sigma(X_{1})$, but I am not sure, as I struggle with the measure theoretical understanding of what an independent copy actually is.