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 $M^2(-\mu^2)=0$ ?
