The number of elements in a finite field must be a prime power, and for every prime power there is only one finite field *up to isomorphism*.

The finite field with 256 elements, GF(2^{8}), is important in applications. From one perspective, there is only one such field. But there are a lot of different isomorphic representations of that field, and some are more efficient to work with that others.

Just how many ways are there to represent GF(2^{8})? Someone with the handle joriki gave a clear answer on Stack Exchange:

There are 2^{8}−1 different non-zero vectors that you can map the first basis vector to. Then there are 2^{8}−2 different vectors that are linearly independent of that vector that you can map the second basis vector to, and so on. In step *k*, 2^{k-1} vectors are linear combinations of the basis vectors you already have, so 2^{8}−2^{k−1} aren’t, so the total number of different automorphisms is

This argument can be extended to count the number of automorphism of any finite field.

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