If $X$ and $Y$ are iid, is $E(X\mid X+Y)=E(Y\mid X+Y)$? Just like question asked, my thought is they have the same mapping.
My classmate gives me a counterexample: if $X$ and $Y$ are equal, then $X+Y=2X$. However, $\operatorname{Var}(2X) \neq \operatorname{Var}(X+Y)$. So $2X$ and $X+Y$ don't have the same distribution? I am confused.
 A: If $X$ and $Y$ are exactly equal, the only way for them to be independent is if they're constant (i.e. have 0 variance). Hence we'd actually have $\text{Var}(2X) = \text{Var}(X + Y) = 0$ in this case.
Like Sangchul Lee said, the way to prove your claim is to note that $(X ,Y)$ and $(Y, X)$ have the same joint distribution.
A: iid: Independent and Identical Distribution. 
If you roll two identical dice, you don't get two identical number($2X$) every time.
$2X$ is not the same as $X+Y$(Sum of two dice). Not even in distribution.    
$2X$ will be either 2,4,6,8,10,12, and the probabilities are all the same $\frac{1}{6}$.
However, $X+Y$ will be one of  2,3,4,5,6,7,8,9,10,11,12 and the probabilities are all different.
Pr($X+Y=12$) = $\frac{1}{36}$ but Pr($X+Y=6$) = $\frac{5}{36}$
Indeed this example is famous as an example of the central limit theorem. 
Therefore, Var($2X$) $\ne$ Var($X+Y$) is what it is supposed to be. 
In fact, Let Var($X$) = $\sigma^2$, then Var($2X$) = 4$\sigma^2$, and Var($X+Y$) = 2$\sigma^2$.
Since we don't distinguish these two dice (identical),  ${E}(X|X+Y) ={E}(Y|X+Y) $
