Compute $E[X_1|Y]$

Let $V = \max\{X_2,\ldots,X_n\}$. Clearly $0\leqslant V \leqslant 1$ with probability 1. For $0\leqslant v\leqslant 1$ we have $$\begin{eqnarray} F_V(v) &=& \Pr(V \leqslant v) = \Pr(\max\{X_2,\ldots,X_n\} \leqslant v) \\ &=& \Pr(X_2 \leqslant v,\ldots,X_n \leqslant v) \\ &=& F_X(v)^{n-1} = v^{n-1} \end{eqnarray} $$ And therefore $f_V(v) = (n-1) v^{n-2} [ 0<v<1]$. Now to the heart of the question: $$\begin{eqnarray} \mathsf{E}(X_1\mid Y=y) &=& \mathsf{E}(X_1\mid \max\{X_1,V\}=y) \\ &=& \mathsf{E}(X_1\mid \max\{X_1,V\}=y,X_1 \leqslant V) \Pr(X_1 \leqslant V) + \\ && \mathsf{E}(X_1\mid \max\{X_1,V\}=y,X_1 > V) \Pr(X_1>V) \\ &=& \mathsf{E}(X_1\mid V=y, X_1\leqslant V) \Pr(X_1 \leqslant V) + \mathsf{E}(X_1\mid X_1=y,X_1>V) \Pr(X_1>V) \\ &=& \frac{y}{2} \Pr(X_1 \leqslant V) + y \Pr(X_1 >V) = \frac{y}{2} \frac{n-1}{n} + y \frac{1}{n} = \frac{y}{2} \frac{n+1}{n} \end{eqnarray} $$ Therefore $$ E(X_{1}|Y)=\frac{Y}{2} \frac{n+1}{n} $$

Added: Here is a verification of this result with Mathematica:

In[17]:= Assuming[n > 1 && 0 < y1 < y2 < 1, 
 Expectation[
   x1 \[Conditioned] 
    y1 < Max[x1, v] < y2, {x1 \[Distributed] UniformDistribution[], 
    v \[Distributed] 
     OrderDistribution[{UniformDistribution[], n - 1}, n - 1]}] // 
  FullSimplify]

Out[17]= (y1^(1 + n) - y2^(1 + n))/(2 y1^n - 2 y2^n)

In[18]:= Limit[%, y1 -> y2, Direction -> 1, 
 Assumptions -> 0 < y2 < 1 && n > 1]

Out[18]= ((1 + n) y2)/(2 n)