$f(f(z)) = z , f(\exp(z)) = \exp(f(z)) $?

Question

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 convergence. 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.

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Answer 1

I don't know the complete answer, but here are some examples:

More generally, let $x$ be a repelling periodic point of period $p$ for an entire map $g$. Then there exists a linearizing coordinate $\varphi$ for $x$; i.e., a conformal map defined near $x$ such that $\varphi(x)=0$ and $\varphi (g^{\circ p}(z)) = \lambda \cdot \varphi(z)$ where $\lambda = (g^{\circ p})' (x)$ be the multiplier of $x$.

Hence, for any primitive $k$-th root $\omega$, $f(z) = \varphi^{-1}(\omega\cdot \varphi(z))$ satisfies $f^{\circ k}(z)=z$ and \begin{align*} f(g^p(z)) &= \varphi^{-1}(\omega\cdot \varphi(g^p(z))) = \varphi^{-1}(\lambda\omega \cdot \varphi(z))) = g^p(\varphi^{-1}(\omega \cdot \varphi(z))) \\ &= g^p(f(z)). \end{align*}

Therefore, the cases $g = \exp$, $k=p=1,2$ give non-trivial examples for (A) and (B) respectively. Note that $J(\exp)=\mathbb{C}$ and all periodic points are repelling.