Why do we have freedom to choose a renormalization scale in massless QFT theories? I thought that the freedom to choose renormalization conditions arises from the freedom to choose the arbitrary renormalization parameters. Let me exemplify this in a
Massive $\phi^4$ scalar theory
Let's denote with a subscript $_0$ bare quantities and let $\phi$ be the renormalized field defined by $\phi_0 =: \sqrt{A} \, \phi$ for some arbitrary constant $A$, so that
\begin{equation}
\begin{split}
\mathcal{L}  &= \frac{1}{2}(\partial \phi_0)^2 - \frac{1}{2}m_0^2\phi_0^2-\frac{1}{4!}\lambda_0\phi_0^4 \\
&\equiv  \frac{1}{2} A(\partial \phi)^2 - \frac{1}{2}m_0^2A\phi^2-\frac{1}{4!}\lambda_0A^2\phi^4 \\
&\equiv \frac{1}{2}(\partial \phi)^2 - \frac{1}{2}\tilde{m}^2\phi^2-\frac{1}{4!}\tilde{\lambda}\phi^4 + \frac{\delta_Z}{2}(\partial \phi)^2 - \frac{\delta_m}{2}\phi^2-\frac{\delta_{\lambda}}{4!}\phi^4 \\
\end{split}
\end{equation}
This is just a rewriting,  identically true for any choice of the renormalization parameters $A, \tilde{m}, \tilde{\lambda}$ upon $$A = 1 + \delta_Z , \quad A m_0^2 = \tilde{m}^2 + \delta_m , \quad A^2 \lambda_0 = \tilde{\lambda} + \delta_{\lambda} $$
Let's focus on $A$ and $\tilde{m}$. From Dyson resummation and Kallen-Lehmann representation the Fourier bare propagator is  $$\mathcal{F}\left[ \langle\Omega|T\phi_0(x)\phi_0(y)|\Omega\rangle \right](p) \equiv G_0(p) = \frac{i}{p^2-m_0^2 - M_0^2(p^2)} = \frac{iZ}{p^2-m^2} + \text{regular at } p^2 = m^2 $$
where $-iM_0^2(p^2)$ is the sum of Fourier amputated 1PI bare diagrams and $m$ is the physical mass.
Then the Fourier renormalized propagator has, on one side, the same Dyson structure of the bare one upon $m_0 \rightarrow \tilde{m}$ and $M_0^2 \rightarrow M^2$, on the other it's simply $\frac{G_0(p)}{A}$ per definition of renormalized field:
$$\mathcal{F}\left[ \langle\Omega|T\phi(x)\phi(y)|\Omega\rangle \right](p) \equiv G(p) = \frac{i}{p^2-\tilde{m}^2 - M^2(p^2)} = \frac{i \, Z/A}{p^2-m^2} + \text{regular at } p^2 = m^2 $$
where $-iM^2(p^2)$ is the sum of Fourier amputated 1PI renormalized diagrams and $m$ is again the physical mass.
In particular the renormalized propagator still has a pole at the physical mass and it's residue there it's $i \, Z/A$:
$$ p^2 - \tilde{m}^2 - M^2(p^2) |_{p^2 = m^2} = 0 \Rightarrow M^2(m^2) = m^2 - \tilde{m}^2 \quad \quad \quad (*) $$
$$ i \, Z/A = \frac{i}{1-\frac{d}{dp^2} M^2(p^2)|_{p^2 = m^2}} \Rightarrow \dot{M}^2(m^2) = 1-(Z/A)^{-1} \quad \quad (**) $$
These are all facts. But we still have the freedom to choose $A$ and $\tilde{m}$: an obvious choice is
$$A \overset{!}{=} Z , \quad \tilde{m} \overset{!}{=}m \Rightarrow M^2(m^2) = 0, \quad \dot{M}^2(m^2) = 0 $$
Another possibile choice is to introduce some renormalization scale $\mu$ so that
$$ M^2(-\mu^2) = 0, \quad \dot{M}^2(-\mu^2) = 0 $$
We can choose 2 conditions because we have 2 free parameters, $A$ and $\tilde{m}$. This choice implicitly constraints these parameters in function of $\mu$ via equations $(*), (**)$; this dependence is described by Callan-Symanzik equation.
Question 1: is my understanding correct so far?

Massless theory
In the massless theory all goes the same way, but since no $\tilde{m}$ is introduced we have only one degree of freedom, $A$, to choose. Following the same reasoning as before without $m_0, m$ and $\tilde{m}$ around the renormalized propagator must still have a pole at $p^2 = 0$, so equation (*) seems constrained to $M^2(0)=0$.
Question 2: Why do we have the freedom to choose again $M^2(-\mu^2)=0$ ?
 A: 
In the massless theory all goes the same way, but since no $\tilde{m}$ is introduced we have only one degree of freedom, $A$, to choose...
  Question 2: Why do we have the freedom to choose again $M^2(-\mu^2)=0$ ?

For massless $\phi^4$ scalar theory, you still have the freedom to choose the mass renormalization condition. The reason is that nothing prevents the scalar field from gaining a mass from quantum corrections. So there is practically no difference between massive and massless $\phi^4$ scalar theories. This is actually all the fuss about surrounding the "naturalness" of the Higgs model: Higgs (as a $\phi^4$ scalar theory) can have mass quantum corrections all the way up to the Planck scale, regardless of the initial mass the Higgs field has. 
There are two cases which can spare you from the huge mass quantum correction predicament:


*

*Local gauge invariance. Photon is massless and will stay that way thanks to the Ward Identity. 

*Global gauge invariance. For example, fermion mass is small because it is protected by the chiral symmetry (a Dirac mass term breaks the chiral symmetry). So an initially small fermion mass will not be spoiled by large quantum corrections. It's dubbed as technical naturalness (a la t'Hooft). But it can not explain why the fermion mass is small in the first place (Dirac's strong naturalness).

