Alright, I feel this is the thread where I should explain why there can't be an iteration about two fixed points--it comes with a few caveats.
To begin, let me define what I refer to as a local iteration, which is a strengthening of regular iteration (so it is more restrictive). Let's assume that \(D\subset \mathbb{C}\) is a domain that is closed under addition of elements. So that \(\{D,+\}\) is a semi-group.
Let's also assume that \(E\) is a SIMPLY CONNECTED DOMAIN (this is the missing criteria which allows for the double fixed point iteration that Bo has found). Let's assume that \(f : E \to E\) and then a local iteration for \(t \in D\) and \(z \in E\), is a holomorphic function:
\[
f^{\circ t}(z) : D \times E \to E\\
\]
Where \(f^{\circ 1}(z) = f(z)\) and this function satisfies \(f^{\circ t_0}(f^{\circ t_1}(z)) = f^{\circ t_0 + t_1}(z)\).
To begin, since \(E\) is simply connected, it is safe to assume that \(E = \mathbb{D}\) the unit disk (conjugate with a riemann mapping). A mapping from the unit disk to itself always has a fixed point, and so again, without loss of generality, we can assume one of the fixed points is \(0\). By schwarz's lemma, this means that \(|f'(0)| \le 1\) and that \(|f(z)| \le |z|\). Where \(|f(z)| = |z|\) only when \(|f'(0)| = 1\). This means that \(f(z) = e^{i\theta}z\) for some \(\theta \in [0,2\pi)\).
So, if \(|f'(0)| < 1\), then \(|f(z)| < |z|\) and there are no more fixed points other than \(0\) by construction. So all that's left is if \(f\) has a neutral fixed point at \(0\). Wlog we can assume that \(f'(0) = 1\). Well then \(f(z) = z\), which is only fixed points, and is therefore the trivial iteration. And can be discarded. Or you can just observe that \(e^{i\theta}z\) has no other fixed points than zero.
So there you have it, this is what I mean when I say you can't have an iteration about two fixed points. Bo's solution explicitly doesn't satisfy these conditions. In fact, you can make infinitely many of these iterations about 2 fixed points.
Let \(h^{\circ t}(z) = \lambda^t z\) for some \(\lambda \in \mathbb{C}/\{0\}\). Take a linear fractional transformation which maps \(\infty \to A\) and \(0 \to B\). Let's call it \(\mu\). Then:
\[
g^{\circ t}(z) = \mu(h^{\circ t}(\mu^{-1}(z)))\\
\]
Is a holomorphic fractional iteration (it will be regular), about both fixed points \(A\) and \(B\).
This clears up what I was looking for, and what I mean by no iteration about two fixed points; I specifically mean no local iteration as I've defined above. Again, I think the main issue we were having, Bo, was semantics.
Hope that makes sense
EDIT: Additionally I should add, that for the case \(b^z\), so long as the fixed point isn't parabolic, then the immediate basin of attraction of any fixed point (or a fixed point of \(\log_b\)), is simply connected. This can be found in Introduction to Chaotic Dynamical Systems by Devaney. By which, the maximal domain of the iteration is this basin, as the Julia set is the boundary of the basin. Additionally any orbit of the julia set gets arbitrarily close to infinity, in a singular manner.
To begin, let me define what I refer to as a local iteration, which is a strengthening of regular iteration (so it is more restrictive). Let's assume that \(D\subset \mathbb{C}\) is a domain that is closed under addition of elements. So that \(\{D,+\}\) is a semi-group.
Let's also assume that \(E\) is a SIMPLY CONNECTED DOMAIN (this is the missing criteria which allows for the double fixed point iteration that Bo has found). Let's assume that \(f : E \to E\) and then a local iteration for \(t \in D\) and \(z \in E\), is a holomorphic function:
\[
f^{\circ t}(z) : D \times E \to E\\
\]
Where \(f^{\circ 1}(z) = f(z)\) and this function satisfies \(f^{\circ t_0}(f^{\circ t_1}(z)) = f^{\circ t_0 + t_1}(z)\).
To begin, since \(E\) is simply connected, it is safe to assume that \(E = \mathbb{D}\) the unit disk (conjugate with a riemann mapping). A mapping from the unit disk to itself always has a fixed point, and so again, without loss of generality, we can assume one of the fixed points is \(0\). By schwarz's lemma, this means that \(|f'(0)| \le 1\) and that \(|f(z)| \le |z|\). Where \(|f(z)| = |z|\) only when \(|f'(0)| = 1\). This means that \(f(z) = e^{i\theta}z\) for some \(\theta \in [0,2\pi)\).
So, if \(|f'(0)| < 1\), then \(|f(z)| < |z|\) and there are no more fixed points other than \(0\) by construction. So all that's left is if \(f\) has a neutral fixed point at \(0\). Wlog we can assume that \(f'(0) = 1\). Well then \(f(z) = z\), which is only fixed points, and is therefore the trivial iteration. And can be discarded. Or you can just observe that \(e^{i\theta}z\) has no other fixed points than zero.
So there you have it, this is what I mean when I say you can't have an iteration about two fixed points. Bo's solution explicitly doesn't satisfy these conditions. In fact, you can make infinitely many of these iterations about 2 fixed points.
Let \(h^{\circ t}(z) = \lambda^t z\) for some \(\lambda \in \mathbb{C}/\{0\}\). Take a linear fractional transformation which maps \(\infty \to A\) and \(0 \to B\). Let's call it \(\mu\). Then:
\[
g^{\circ t}(z) = \mu(h^{\circ t}(\mu^{-1}(z)))\\
\]
Is a holomorphic fractional iteration (it will be regular), about both fixed points \(A\) and \(B\).
This clears up what I was looking for, and what I mean by no iteration about two fixed points; I specifically mean no local iteration as I've defined above. Again, I think the main issue we were having, Bo, was semantics.
Hope that makes sense
EDIT: Additionally I should add, that for the case \(b^z\), so long as the fixed point isn't parabolic, then the immediate basin of attraction of any fixed point (or a fixed point of \(\log_b\)), is simply connected. This can be found in Introduction to Chaotic Dynamical Systems by Devaney. By which, the maximal domain of the iteration is this basin, as the Julia set is the boundary of the basin. Additionally any orbit of the julia set gets arbitrarily close to infinity, in a singular manner.
