Is there anything like GF(6)?

No; finite fields must have order a power of prime, and for every prime $p$ and every $n\gt 0$, there is one and only one (up to isomorphism) field of order $p^n$.

To see why the order must be the power of a prime: note that the characteristic of an integral domain must be a prime; since a field is an integral domain, it must be of characteristic $p$ for some prime $p$. It is now easy to see that the field is in fact a vector space over the field of order $p$; since it is finite, it must have a basis with $n$ elements; but a vector space of dimension $n$ over a field with $p$ elements must have $p^n$ vectors.

To see why there is one and only one up to isomorphism, consider the splitting field of $x^{p^n}-x$ over $GF(p)$.

No, because for any reasonable interpretation of the example given, 2 and 3 are zero-divisors (as is 4) and have no multiplicative inverse, so the structure is not a field.

However there are Galois extensions of fields which have degree 6 (the general cubic over the rationals is an example) - and the Galois group then has order 6.