Consider the following exercise
If $X_1,\dots,X_n$ are independent with same distribution and $S_n=X_1+\cdots+X_n$, prove that$$\mathbb{E}[X_i \mid S_n]=\frac{S_n}{n}$$Proof: Since the variables have the same distribution we have that:$$S_n=\mathbb{E}[S_n \mid S_n]=\mathbb{E}\left[\sum_{i=0}^n X_i \mid S_n \right]=\sum_{i=0}^n\mathbb{E}[X_i \mid S_n]=n \cdot \mathbb{E}[X_i \mid S_n]$$
I'm trying to figure out why the fact that the variables have the same distribution implies that they have the same conditional expectation with respect to $S_n$. I tried to see if the expected values of the variables on the sets of the form $S_n^{-1}(A)$ (with $A\subseteq \mathbb{R}$ borel set) are equal (I mean that $\mathbb{E}[X_1\mathbf{1}_{S_n^{-1}(A)}]=\mathbb{E}[X_i\mathbf{1}_{S_n^{-1}(A)}]$), because this fact would give me the thesis, but I don't know how to do it. Any suggestion on how to demonstrate the equality between the conditional expectations?
