The only working $k$-tuple is $(1,1, \ldots, 1)$. Notice that if $n_i=1$, then $n_{i+1}=1$, so if one of the numbers is $1$, all of them are $1$. Thus, assume all of them are greater than $1$ and let $p_i$ be the minimal prime divisor of $n_i$. By a classical order argument, either $\text{gcd}(p_{i+1}-1, n_i)>1$, or the order is $1$ and $p_{i+1} \mid n_i \implies p_{i+1} \geq p_i$. In the first case, there is a prime $q$, such that $q \mid n_i$ and $q \mid p_{i+1}-1$, so $p_i \leq q \leq p_{i+1}-1$, and thus $p_{i+1}>p_i$. Thus, in both cases, $p_{i+1} \geq p_i$, and applying this cyclically yields the $p_i$ are constant, and thus all $n_i$ are divisible by some fixed prime, contradiction with the fact that they are coprime.