Suppose as an example with $n=3$ we have that $X_1, X_2, \dots$ is a sequence of random variables satisfying the condition that $(X_1,X_2,X_3), (X_4,X_5,X_6), (X_7,X_8,X_9), ...$ and so on are all i.i.d. $3$-tuples. Does this imply that, for example, $(X_2, X_3, X_4)$ and $(X_3, X_4, X_5)$ are i.i.d.?
I certainly know that our condition above doesn't imply that all of the individual $X_i's$ are i.i.d. and could quickly construct a counter-example. However I can't think of a counter-example disprove the possibility of these shifted $3$-tuples being i.i.d. Either a counter-example or some sketch of a proof showing why they're i.i.d. would be greatly appreciated.
I also suspect this has some relation to the concept of stationarity.
No, they are not IID. Recall that if two random vectors $X$ and $Y$ are independent, then so are all Borel images $f(X)$ and $g(Y)$ of them.
Arguing by contradiction, suppose $(X_2, X_3, X_4)$ and $(X_3,X_4,X_5)$ are indepedent. If we take $f = \pi_2$ and $g=\pi_1$ (the projections onto the second and first coordinates, respectively), then you get $X_3 = f(X_2, X_3, X_4) = g(X_3,X_4,X_5)$, meaning that $X_3$ is independent of itself. This cannot happen if $X_3$ is not a constant.
Edit: In case you meant to ask the variant of this question proposed by @MikeEarnest in the comments, the answer is still no. To see this, we let the tuples $\vec{X}_i = (X_{3i-2}, X_{3i-1}, X_{3i})$ by IID as $i$ varies, but chosen so that $X_{3i-2} = X_{3i-1}$. Then, the vectors $(X_2, X_3, X_4)$ and $(X_5, X_6, X_7)$ cannot be independent by considering the projections $\pi_1$ and $\pi_3$ and observing $$\pi_1 (X_5, X_6, X_7) = X_5 = X_4 = \pi_3 (X_2, X_3, X_4)$$ implying the proposed random vectors are not independent.