The field norm and trace exist when $K$ is a finite algebraic extension of $F$.  In this case, an element $\alpha \in K$ can be interpreted as an $F$-linear map on $K$ by multiplication.  The field norm is just the determinant of $\alpha$ as a linear map, while the trace is the trace of $\alpha$ as a linear map.  This yields an evident generalization:  Norm and trace are part of a family of nice maps, namely the coefficients of the characteristic polynomial of $\alpha$.

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Since Zev asks for a uniqueness theorem in the comments, here is one that shows both the merits and limitations of the characteristic polynomial as an answer.

For simplicity let $F$ have characteristic 0.  Let $K$ be a field extension of degree $n$ which is generic in the sense that the Galois group is $S_n$.  Then any Galois-invariant of degree $n$ from $K$ to $F$ which is a polynomial in $\alpha \in K$ and its Galois conjugates, is a symmetric polynomial.  The theorem is that the algebra of symmetric polynomials is generated by elementary symmetric polynomials, which are exactly the coefficients of the characteristic polynomial of $\alpha$.  (This is using the fact that the eigenvalues of $\alpha$ as a map are itself and its Galois conjugates.)  In particular, the trace is the unique linear such map up to a scalar; and any multiplicative polynomial of this type is a power of the norm.  You can also describe the norm as the last Galois-invariant polynomial (the one of degree $n$) that provides new information.

But if the Galois group is smaller, then the ring of invariant polynomials in $\alpha$ and its Galois conjugates is larger, and any of these other invariant polynomials is also "nice".

Well, the original question was open-ended.  I think that this answer does fit one interpretation of the question, but maybe it is too standard and maybe there are also other interesting answers.