Given $f$ holomorphic, which are the necessary conditions on $\phi$ in order to make $\phi \circ f \circ \phi^{-1}$ holomorphic? It is well known that $\bar f(\bar z)$ is holomorphic whenever f is. I was wondering how to generalize this fact...
Let $f: \Omega \longrightarrow \mathbb{C}$ be holomorphic and $\phi: \mathbb{C} \longrightarrow \mathbb{C}$ be an homeomorphism where $\Omega \subseteq \mathbb{C}$ is open.
We need the exsistence of the limit $\lim_{h \rightarrow 0} \frac{\phi \circ f \circ \phi^{-1}(z_0 + h) - \phi \circ f \circ \phi^{-1}(z_0)}{h}$, if $\phi$ is Frechet differentiable this is equivalent to asking for the existence of $\lim_{h \rightarrow 0} \frac{D\phi(f \circ \phi^{-1}(z_0))[f'(\phi^{-1}(z_0))\cdot D\phi^{-1}(z_0)[h]]}{h}$.
I've then found the following sufficient conditions:
i) $\phi(z + w) = \phi(z) + \eta(w)$
ii) $\eta(z \cdot w) = \psi(z) \cdot \eta(w)$
Where $\eta,\psi: \mathbb{C} \longrightarrow \mathbb{C}$ and $\eta$ is an homeomorphism.
Then $\forall z_0 \in \mathbb{C}.$ $D\phi(z_0)$ exists and $D\phi(z_0) = \eta$ thus $\forall y_0 \in \mathbb{C}$. $D\phi^{-1}(y_0) = \eta^{-1}$.
Moreover we have $D\phi(f \circ \phi^{-1}(z_0))[f'(\phi^{-1}(z_0))\cdot D\phi^{-1}(z_0)[h]] = \eta(f'(\phi^{-1}(z_0)) \cdot \eta^{-1}(h)) = \psi(f'(\phi^{-1}(z_0))) \cdot h$, thus the limit exists and has value $\psi(f'(\phi^{-1}(z_0)))$
As an example we can take $\phi(z) = \alpha z + \beta$ with $\alpha, \beta \in \mathbb(C)$, then $\eta(z) = \alpha z$ and $\psi(z) = z$ thus $(\phi \circ f \circ \phi^{-1})'(z_0) = f'(\frac{z_0}{\alpha} - \frac{\beta}{\alpha})$ and sure enough if we use the standard method to evaluate this derivative we get the same resut.
Conditions i) and ii) above are then sufficient, are they necessary too? If not does there exist a complete characterization of such $\phi$'s?
 A: This is actually a question about topological group rather than complex analysis. I will extend $\phi$ as in your question to a homeomorphism of the Riemann sphere, by sending $\infty$ to itself. 
Every homeomorphism $\phi$ as in your question defines an automorphism $\phi_*$ of the semigroup $Hol$ of holomorphic maps $f: {\mathbb C}\to {\mathbb C}$,
$$
\phi_*(f)= \phi \circ f \circ \phi^{-1}. 
$$ 
Hence, it preserves the subgroup $A$ of $Hol$ consisting of invertible elements. This subgroup consists of complex-affine maps 
$$
z\mapsto az+b, a\in {\mathbb C}^*, b\in {\mathbb C}. 
$$ 
The automorphism $\phi_*$ of $A$ is continuous (in the standard topology on $A$). 
The group $A$ contains infinite cyclic subgroups $C$ generated by Euclidean rotations
$$
z\mapsto az+b, |a|=1,  
$$
$a$ is not a root of unity. By continuity, the automorphism $\phi_*$ has to send such subgroups to infinite cyclic subgroups of Euclidean rotations.  Orbits of every such subgroup $C$ are dense subsets of circles in ${\mathbb C}$. Conversely, for every circle in ${\mathbb C}$ appears as the the orbit closure of one of such subgroups $C< A$. Therefore, by continuity, $f$ maps Euclidean circles to Euclidean circles. 
A similar argument works for Euclidean lines: $f$ sends real affine lines to lines. One can either prove this by observing that every line is the limit of a sequence of expanding circles or by noting that for every affine line $L\subset {\mathbb C}$ there exists a subgroup $H$ of translations in $A$ isomorphic to ${\mathbb Z}^2$, preserving $L$ and acting on $L$ such that every $H$-orbit in $L$ is dense. 
From this, similar to circles, one concludes  that $\phi$ maps lines to lines. 
Now, one uses a classical fact that every homeomorphism of the Riemann sphere sending circles to circles (circles passing through infinity are affine lines $\cup \{\infty\}$) is a Moebius transformation, i.e. it has the form
$$
z\mapsto \frac{az+b}{cz+d}
$$ 
or
$$
z\mapsto \frac{a\bar{z}+b}{c\bar{z}+d}. 
$$
This is easy to prove, see e.g. J.Andersen's book "Hyperbolic Geometry."  (It was discussed at MSE earlier, here.) 
Our map $\phi$ fixes the point $\infty$, hence, it has the form either of 
$$
z\mapsto a{z}+b. 
$$ 
or 
$$
z\mapsto a\bar{z}+b. 
$$ 
A: This is only a partial answer under the additional assumption that $\phi$ is a diffeomorphism. While I believe that your assumptions do imply that $\phi$ is a diffeomorphism, this is not obvious and I do not have a proof at this moment.
First assume that $\phi$ fixes zero, so $\phi(z) = az + b \bar{z} + o(|z|)$ near zero, where $a = \frac{\partial \phi}{\partial z}(0)$ and $b = \frac{\partial \phi}{\partial \bar{z}}(0)$ are the complex (or Wirtinger) partial derivatives at zero. Then for $f(z)=iz$ we have that $g = \phi \circ f \circ \phi^{-1}$ is holomorphic and fixes zero, so $g(z) = cz+o(|z|)$ near zero. Writing out the linear parts of the equation $g \circ \phi = \phi \circ f$ at zero, we get $iaz-ib\bar{z} = caz+cb\bar{z}$, so that $ia=ca$ and $-ib=cb$. Since $c$ can not be equal to both $i$ and $-i$, this shows that $a=0$ or $b=0$, i.e., $\frac{\partial \phi}{\partial \bar{z}}(0)=0$ or $\frac{\partial \phi}{\partial {z}}(0)=0$.
Either using translations, or repeating the argument for the case $\phi(z_0)=w_0$, where $z_0$ and $w_0$ are arbitrary points in the plane, one gets that $\frac{\partial \phi}{\partial \bar{z}}(z_0)=0$ or $\frac{\partial \phi}{\partial {z}}(z_0)=0$ for all $z_0$. Since we assumed that $\phi$ is a diffeomorphism, these derivatives can not simultaneously vanish, and a connectedness argument shows that either $\frac{\partial \phi}{\partial \bar{z}}(z_0)=0$ for all $z_0$, or $\frac{\partial \phi}{\partial {z}}(z_0)=0$ for all $z_0$. In the first case, $\phi$ satisfies the Cauchy-Riemann equations, so it is an analytic diffeomorphism of the plane, which means that $\phi(z)=\alpha z + \beta$ with $\alpha \ne 0$. In the second case, $\phi$ is an analytic diffeomorphism of $\bar{z}$, so $\phi(z) = \alpha \bar{z} + \beta$.
