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If we have i.i.d. random variables, $X$ and $Y$, then $\text{Cov}(X,Y)=0$.

But let's say we have i.i.d. random vectors $\boldsymbol{X}$ and $\boldsymbol{Y}$, where $\boldsymbol{X}=(X_{1},...,X_{p})$ and $\boldsymbol{Y}=(Y_{1},...,Y_{q})$. Do we have any properties analogous to the univariate case: $\text{Cov}(X,Y)=0$, that always hold true, as a result of these random vectors being i.i.d.? Something like $\text{Cov}(X_i,X_j)=0$ if $i\neq j$, or $\text{Cov}(X_i,Y_i)=0$, or maybe the independence of $\boldsymbol{X}$ and $\boldsymbol{Y}$ gives a covariance matrix with zero terms everywhere except the diagonal (these are all just guesses, and not necessarily true). Are there any properties like this for i.i.d. random vectors?

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If the random vectors $X, Y$ are independent, then, as functions of independent random variables are independent (see Functions of Independent Random Variables), $X_i$ and $Y_j$ are independent, since selecting one given component from a vector is a function. But the covariance of independent random variables are zero, that gives you the result.

In your question you speculate about a condition $i \not= j$, but that is unnecessary here. $X_i$ and $Y_i$ re also independent. The covariance matrix $\DeclareMathOperator{\C}{\mathbb{C}} \C(x,y)$ will be the zero matrix, also the diagonal.

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