Consider the random variables $\{X,Y\}_{i=1}^n$ defined on the same probability space $(\Omega, \mathcal{F}, P)$. $\{X_i\}_{i=1}^n$ are i.i.d. across $i$. $\{Y_i\}_{i=1}^n$ are i.i.d. across $i$ conditional on $X_1,...,X_n$.
Let $\mathcal{N}:=\{1,2,...,N\}$ and consider $\varphi:\mathcal{N}\rightarrow \mathcal{N}$ being a bijective function representing a permutation of labels.
Supposing for simplicity that the random variables are discrete and $n=4$, is it true that
(*) $$ P(Y_1=1,Y_2=2, Y_3=3, Y_4=4|X_1=a, X_2=b, X_3=c, X_4=d)=P(Y_{\varphi(1)}=1,Y_{\varphi(2)}=2, Y_{\varphi(3)}=3, Y_{\varphi(4)}=4|X_{\varphi(1)}=a, X_{\varphi(2)}=b, X_{\varphi(3)}=c, X_{\varphi(4)}=d) $$ ? I believe the answer is yes because
(i) using $\{Y_i\}_{i=1}^n$ i.i.d. across $i$ conditional on $X_1,...,X_n$ $$ P(Y_1=1,Y_2=2, Y_3=3, Y_4=4|X_1=a, X_2=b, X_3=c, X_4=d)=P(Y_{\varphi(1)}=1,Y_{\varphi(2)}=2, Y_{\varphi(3)}=3, Y_{\varphi(4)}=4|X_1=a, X_2=b, X_3=c, X_4=d) $$
(ii) Using $\{X_i\}_{i=1}^n$ i.i.d. across $i$ $$ P(Y_{\varphi(1)}=1,Y_{\varphi(2)}=2, Y_{\varphi(3)}=3, Y_{\varphi(4)}=4|X_1=a, X_2=b, X_3=c, X_4=d)=P(Y_{\varphi(1)}=1,Y_{\varphi(2)}=2, Y_{\varphi(3)}=3, Y_{\varphi(4)}=4|X_{\varphi(1)}=a, X_{\varphi(2)}=b, X_{\varphi(3)}=c, X_{\varphi(4)}=d) $$
Correct?