Let $\{X_i\}$ be a sequence of (not necessarily independent) identically distributed random variables defined on a common probability space. What conditions do we need to impose on $\{X_i\}$ so that for any $n \in \mathbb{N}$, the distributions of the random variables $\{Y_i\}_{i=1}^{n}$ with $Y_i = (X_1,\cdots,X_{i-1},X_{i+1},\cdots,X_n)$ are the same? What about generalizations to leave-K-out versions where we consider the $N := {n \choose k}$ random variables $\{Y_i\}_{i=1}^{N}$ by leaving out any $k$ of the $X_j$'s?
My thoughts: Clearly, a sufficient condition is independence of $\{X_i\}$ (or more generally, exchangeability); however, I am hopeful that this holds under much weaker conditions. One property that pops into my mind is stationarity of the sequence $\{X_i\}$, but I am not sure if this is sufficient since the definition of stationarity requires invariance of the joint distribution to shifting and not rearrangements (it seems like stationarity might be necessary since we want $(X_1,\cdots,X_{n-1})$ and $(X_2,\cdots,X_n)$ to have the same distribution for any $n$).
EDIT: I think stationarity is sufficient for the leave-one-out case, and I'd appreciate it if someone can verify the below argument.
Suppose $n \in \mathbb{N}$, and consider the leave-one-out sequence $\{Y_i\}_{i=1}^{n}$ defined above. We want to verify that the distribution function of $Y_j$ is identical to that of $Y_k$ whenever $j,k\in \{1,\cdots,n\}$, and $j < k$. Consider the subsequence $\{X_i\}_{i=j}^{k}$ and note that by stationarity of $\{X_i\}$, the distribution functions of $\{X_i\}_{i=j}^{k-1}$ and $\{X_i\}_{i=j+1}^{k}$ are identical. That is, conditional on the values of $X_1,\cdots,X_{j-1},X_{k+1},\cdots,X_n$, the distribution functions of $Y_j$ and $Y_k$ are the same. Therefore, the distribution functions of $Y_j$ and $Y_k$ are the same.