Iterated AES encryption under a fixed key
Under a fixed key, $E_K$ is a permutation in the symmetric group of permutations on $2^{128}$ elements. Thus its order divides $(2^{128})!$.

If AES is modelled as a random permutation, for a randomly chosen fixed key, the order $N$ of $E_K$ is likely to be near $2^{127}$ as explained in the comment under the question, and indeed $E_K^{N-1}(x)=E_K^{-1}(x)$ for all $x\{0,1\}^{128}$.