Simple proof that shows AES is not a uniform permutation on any n-bit string? Is there a simple proof that shows AES is not a uniform permutation on any $n$-bit string?

Since I'm just starting with crypto, I'd like to see a simple yet elegant proof for the said property. Thanks!
There is no uniform permutation; there is a permutation uniformly chosen from the set of all possible permutations over $Z_2^{128}$.

It is evident that AES is not a uniformly chosen permutation, since its permutation is fixed for any key.

One can consider a family $\{AES_K\}$  of AES permutations under all possible keys $K$. Even if the key is chosen uniformly, the resulting permutation is not uniformly chosen, as not every permutation is an AES permutation with some key. This comes from a simple counting argument: there are $2^{128}$ 128-bit keys and thus $2^{128}$ AES-128 permutations, but the total number of bijections over $Z_2^{128}$ is
$$
(2^{128})! \approx \frac{2^{128\cdot  2^{128}}}{e^{2^{128}}} \approx 2^{2^{133.5}}.
$$