Consider i.i.d random variables $X_1,X_2,\ldots,X_n$ with an absolutely continuous distribution. Suppose $\boldsymbol Y=(X_{(1)},X_{(2)},\ldots,X_{(n)})$ is the full set of order statistics corresponding to $\boldsymbol X=(X_1,X_2,\ldots,X_n)$.
If $\mathcal P_n$ is the set of $n!$ permutations of $(1,2,\ldots,n)$, then the conditional distribution of $\boldsymbol X$ given $\boldsymbol Y$ is known to be uniform over $\mathcal P_n$. So if $T_n(\boldsymbol X)$ is any function of $\boldsymbol X$, we can say
$$E\left[T_n(\boldsymbol X)\mid \boldsymbol Y\right]=\frac1{n!}\sum_{(i_1,i_2,\ldots,i_n)\in \mathcal P_n}T_n(X_{i_1},X_{i_2},\ldots,X_{i_n}) \tag{$\star$}$$
But does $(\star)$ hold in much more generality, i.e. without the absolute continuity assumption? If not, what can we say about $E\left[T_n(\boldsymbol X)\mid \boldsymbol Y\right]$ if we only know that $X_1,\ldots,X_n$ are i.i.d random variables? We can assume $T_n$ has finite mean and variance.