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I don't know if this actually counts (since I don't know if this functional equation is actually useful...), but in the paper

R.M. Dicker, A set of independent axioms for a field and a condition for a group to be the multiplicative group of a field, Proc. London Math. Soc., 18, 1968, p. 114-124

you can find the following theorem:

Let $A$ be an abelian group (written multiplicatively). Adjoin a new element $0$ to get the set $F$. Assume that there is a function $f : F \to F$ such that for all $x,y \in F$ with $y \neq 0$:

1) $f(0) = 1$

2) $f(f(x))=x$

3) $f(f(x) f(y)) = y f(x f(y^{-1}))$

Then $F$ can be made into a field such that $A = F^*$ and $f(x) = 1 - x$.

Here is another example from algebra which can be found in the article

Zoran Sunik, An Ideal Functional Equation with a Ring, Mathematics Magazine 77 (2004) 4, 310--313.

If $R$ is an integral domain, then the maps $f : R \to R$ solving the functional equation

$f(xz - y) f(x) f(y) + 3 f(0) = 1 + 2 f(0)^2 + f(x) f(y)$

are exactly the characteristic functions of ideals of $R$.

Also functional equations describing subrings and prime ideals are mentioned there.

2 added 27 characters in body

I don't know if this actually counts (since I don't know if this functional equation is actually useful...), but in the paper

R.M. Dicker, A set of independent axioms for a field and a condition for a group to be the multiplicative group of a field, Proc. London Math. Soc., 18, 1968, p. 114-124

you can find the following theorem:

Let $A$ be an abelian group (written multiplicatively). Adjoin a new element $0$ to get the set $F$. Assume that there is a function $f : F \to F$ such that:

1) $f(0) = 1$

2) $f(f(x))=x$

3) $f(f(x) f(y)) = y f(x f(y^{-1}))$

Then $F$ can be made into a field such that $A = F^*$ and $f(x) = 1 - x$.

1

I don't know if this actually counts (since I don't know if this functional equation is actually useful...), but in the paper

R.M. Dicker, A set of independent axioms for a field and a condition for a group to be the multiplicative group of a field, Proc. London Math. Soc., 18, 1968, p. 114-124

you can find the following theorem:

Let $A$ be an abelian group. Adjoin a new element $0$ to get the set $F$. Assume that there is a function $f : F \to F$ such that:

1) $f(0) = 1$

2) $f(f(x))=x$

3) $f(f(x) f(y)) = y f(x f(y^{-1}))$

Then $F$ can be made into a field such that $A = F^*$ and $f(x) = 1 - x$.