While talking about tetration with my friend the following idea (re)occured.
$$f(f(z)) = z ,\quad f(\exp(z)) = \exp(f(z)) \tag{A}\label{A}$$
or variations of it like the weaker
$$f(f(f(f(z)))) = z ,\quad f(\exp(\exp(z))) = \exp(\exp(f(z))). \tag{B}\label{B}$$
Now clearly if $f$ satisfies equation \ref{A} it also satisfies equation \ref{B}, but not necessarily the other way around.
What are analytic solutions $f(z)$ for some local region $D$ bounded by a jordan curve $C$ where $z,f(z),\exp(z),f(\exp(z)),\exp(f(z))$ are in $D$ ( $D$ being the interior part of the curve $C$), that satisfy
$$f(f(z)) = z,\quad f(\exp(z)) = \exp(f(z))?$$
without the trivial $f(z)=z$ ofcourse.
edit
One idea is to use Carleman matrices and Diagonalize.
Consider the Carleman matrix of size $n \times n$ for the exp :
$$M_n(\exp(z)) =A_n$$.
Now diagonalize
$$A_n = P_n \times D_n \times P_n^{-1}$$
Then we can define $f(z)$ by Carleman matrix of size $n \times n$ :
$$M_n(f(z)) =B_n$$.
$$B_n = P_n \times D*_n \times P_n^{-1}$$
where $D*_n$ is the diagonal matrix of size $n$ with all $-1$ on the diagonal.
This way we get
$$B_n^2 = D°$$
where $D°$ is the (diagonal) Carleman matrix of $id(z)$.
and
$$A_n B_n = B_n A_n$$
as desired.
Now let $n$ increase and hope for convergeanceconvergence. And nonzero radius.
And ofcourse we assume diagonalization is always possible.
Might this work ? How to prove it ? What is the result ?
This method does seem to prefer the carleman interpretation of tetration ofcourse, which is not my bias but a consequence of the idea of using eigenvalues and matrices by lack of other " simple " ideas.
