Given $f$ holomorphic, which are the necessary conditions on $\phi$ in order to make $\phi \circ f \circ \phi^{-1}$ holomorphic?

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 $\phi(z) = \alpha \bar{z} + \beta$.

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 $$ z\mapsto a\bar{z}+b. $$