One can verify that also by direct calculation.
Assume we have these fixed points \(z_0\) and \(z_\infty\) where the Kœnigs/Schröder function f is 0 and \(\infty\) (has a pole) respectively.
Let \(\ell_c(z)=cz\) and \(\nu(z)=1/z\), then we take the derivative:
\begin{align}
(f^{-1}\circ \ell_c \circ f)' &= \frac{1}{f'\circ f^{-1}\circ \ell_c\circ f}\cdot (\ell_c'\circ f)\cdot f'\\
(f^{-1}\circ \ell_c \circ f)'(z_0) &= \frac{1}{f'(z_0)} c f'(z_0) = c
\end{align}
However for \(z_\infty\) we can not use this derivation because \(f'(z_0)=\infty\)? We need to find a safer way. But we found already the trick to take the inverse of \(f\) which is analytic at \(z_\infty\). So let \(f=\nu\circ g\) then
\[f^{-1}\circ \ell_c \circ f = g^{-1} \circ \nu \circ \ell_c \circ \nu\circ g = g^{-1}\circ \ell_{1/c} \circ g\]
and as above:
\[(f^{-1}\circ \ell_c \circ f )'(z_\infty) = (g^{-1}\circ \ell_{1/c} \circ g)'(z_\infty) = \frac{1}{c} \]
This is amazing in the sense that as long as the Kœnigs/Schröder function has an analytic 0 and a pole then the corresponding t-iterations have \(c^t\) and \(c^{-t}\) as fixed point derivations. So one could assume that this is the normal situation. However we consider it rather the exception, why?
Because the regular Kœnigs/Schröder function at one fixed point is often not analytic/a pole at the other fixed point.
Though we could have \(\lim_{z\to z_\infty} f^{-1}(c^t f(z)) = z_\infty \) for all t, the limit \(\lim_{z\to z_\infty} (f^{-1}\circ \ell_{c^t} \circ f)'(z) \) would be some other number for discrete t for those functions.
Quite mysterious I would say.
On the other hand if the fixed point derivations are not reciprocal, like with \(\exp_{\sqrt{2}}'(2) = \log(2)\) and \(\exp_{\sqrt{2}}'(4) = 2\log(2) \) then we can surely say that the regular iteration at one fixed point can not be the regular iteration at the other for all t (or just a continuous interval) because as Tommy pointed out: the superfunctions have different periods then.
Assume we have these fixed points \(z_0\) and \(z_\infty\) where the Kœnigs/Schröder function f is 0 and \(\infty\) (has a pole) respectively.
Let \(\ell_c(z)=cz\) and \(\nu(z)=1/z\), then we take the derivative:
\begin{align}
(f^{-1}\circ \ell_c \circ f)' &= \frac{1}{f'\circ f^{-1}\circ \ell_c\circ f}\cdot (\ell_c'\circ f)\cdot f'\\
(f^{-1}\circ \ell_c \circ f)'(z_0) &= \frac{1}{f'(z_0)} c f'(z_0) = c
\end{align}
However for \(z_\infty\) we can not use this derivation because \(f'(z_0)=\infty\)? We need to find a safer way. But we found already the trick to take the inverse of \(f\) which is analytic at \(z_\infty\). So let \(f=\nu\circ g\) then
\[f^{-1}\circ \ell_c \circ f = g^{-1} \circ \nu \circ \ell_c \circ \nu\circ g = g^{-1}\circ \ell_{1/c} \circ g\]
and as above:
\[(f^{-1}\circ \ell_c \circ f )'(z_\infty) = (g^{-1}\circ \ell_{1/c} \circ g)'(z_\infty) = \frac{1}{c} \]
This is amazing in the sense that as long as the Kœnigs/Schröder function has an analytic 0 and a pole then the corresponding t-iterations have \(c^t\) and \(c^{-t}\) as fixed point derivations. So one could assume that this is the normal situation. However we consider it rather the exception, why?
Because the regular Kœnigs/Schröder function at one fixed point is often not analytic/a pole at the other fixed point.
Though we could have \(\lim_{z\to z_\infty} f^{-1}(c^t f(z)) = z_\infty \) for all t, the limit \(\lim_{z\to z_\infty} (f^{-1}\circ \ell_{c^t} \circ f)'(z) \) would be some other number for discrete t for those functions.
Quite mysterious I would say.
On the other hand if the fixed point derivations are not reciprocal, like with \(\exp_{\sqrt{2}}'(2) = \log(2)\) and \(\exp_{\sqrt{2}}'(4) = 2\log(2) \) then we can surely say that the regular iteration at one fixed point can not be the regular iteration at the other for all t (or just a continuous interval) because as Tommy pointed out: the superfunctions have different periods then.
