This question is a sequel to my question Is it possible to produce a field with three operations?.
Using the same definition of/notation for fields, is it possible to have $(F, +, \cdot, *, \dagger)$ such that all of $(F, +, \cdot)$, $(F \setminus \{0\},\cdot, *)$ and $(F \setminus \{0,1\},*, \dagger)$ are fields?
What about if we wanted to chain more operations on?
The answer to the first question, as helpfully pointed out by @JasonDeVito in a comment to my original question, is yes.
To preserve and expand upon it: The field with four elements, $\mathbb{F}_4 = \{0, 1, \alpha, 1+\alpha\}$, can support the structure of $(F, +, \cdot, *, \dagger)$, with $+$ and $\cdot$ defined in the usual way for $\mathbb{F}_4$; with $*$ as the operation defined by @lonzaleggiera, with $\alpha$ defined as the $*$-identity; and with $\dagger$ as the operation defined by @DavidPement, with the somewhat dubious convention that $0 < 1 < \alpha < 1+\alpha$.
It is also the only such field (up to isomorphism). Why? Because both $(F, +, \cdot, *)$ and $(F \setminus \{0\}, \cdot, *, \dagger)$ must be fields with three operations. By the classification collectively provided by the answers to my previous question, this means that, either both $|F|$ and $|F \setminus \{0\}|$ are Mersenne powers of $2$ (that is, powers of $2$ one greater than a prime), or one of the two sizes is $3$ and the other is a Mersenne power of $2$. Since $|F| = |F \setminus \{0\}| +1$, and there are no consecutive Mersenne powers of $2$, the only option is $|F| = 4$ and $|F \setminus \{0\}| = 3$, which, by the classification collectively provided, also forces the operations to be as defined (although $\alpha$ and $1+\alpha$ could be switched, provided it is done for both $*$ and $\dagger$, but this is isomorphic).
As for the second question, the answer is no, and the proof is straightforward:
Suppose $(F, +, \cdot, *, \dagger, \#, ...)$ is a field with at least five operations.
This implies $(F, +, \cdot, *, \dagger)$ and $(F \setminus \{0\}, \cdot, *, \dagger, \#)$ must both be fields with four operations, which implies $|F| = |F \setminus \{0\}| = 4$. But $|F| = |F \setminus \{0\}|+1 = 4+1 = 5$, a contradiction.
Thus, there is no field with five or more operations, QED.