Let $X_1,..., X_n$ a random sample. Let $T=X_1$ and $U=X_1+...+X_n$. The followings is correct?
$\mathbb{E}\left[\left.T\right|U=u\right]=\mathbb{E}\left[\left.X_{1}\right|X_1+...+X_n=u\right]=1/n*\mathbb{E}\left[\left.X_1+...+X_n\right|X_1+...+X_n=u\right]=1/n*u$
Is the second equality correct? why?
The second quality is correct. That's because the $X_i$ are i.i.d, so you can for example exchange $X_1$ with $X_2$ and get again random variables that are i.i.d with the same distribution. That means
$$\mathbb E[X_1|X_1+X_2+\ldots+X_n=u] = \mathbb E[X_2|X_2+X_1+\ldots+X_n=u].$$
I other words, which random variable is "first" and named "$X_1$" is arbitrary if they are all i.i.d. Of course, the sum in the conditional is the same in both cases, so if you generalize this to exchanging $X_1$ with any $X_i$, you get
$$\forall i:1 \le i \le n: \mathbb E[X_1|X_1+\ldots+X_n=u] = \mathbb E[X_i|X_1+\ldots+X_n=u].$$
If you sum this up over all $i$, then apply linearity of expectation you get
$$n\mathbb E[X_1|X_1+\ldots+X_n=u] = \mathbb E[X_1+\ldots+X_n|X_1+\ldots+X_n=u]$$
The second equality you were asking about is now just the the above, divded by $n$.