When is a permutation k-wise independent?
A permutation cannot be $k$-wise independent. A distribution of permutations is $k$-wise independent if for $\pi$ sampled according to the distribution and any $k$ different indices $i_1,\ldots,i_k$ and $k$ different values $j_1,\ldots,j_k$, the probability that $\pi(i_1)=j_1,\ldots,\pi(i_k)=j_k$ is $1/n(n-1)\cdots(n-k+1)$. (An earlier version of this answer contained a different definition that, while interesting, is not the standard one.)

A $k$-wise independent distribution is $\ell$-wise independent for any $\ell \leq k$, and the uniform distribution over all permutations is trivially $n$-wise independent. A $k$-wise independent distribution is a distribution which "looks like" the uniform distribution when considering at most $k$ points at a time.

It is easy to see that any $k$-wise independent distribution needs to have support $n^{\Omega(k)}$, and conversely, Alon and Lovett construct such distributions with support $n^{O(k)}$.

Here is a simple example of a $1$-wise independent distribution: it is the uniform distribution over the permutations $\pi_1,\ldots,\pi_n$ given by $\pi_t(x) = (x+t) \pmod{n}$. It is easy to check that for all $1 \leq i,j \leq n$, the probability that $\pi_t(i) = j$ is exactly $1/n$.