= Solution
Let $\phi_B=Z_\phi^{1/2}\phi$ define the renormalized field in <Phi-fourth theory>. For an $n$-point <correlation function>, with no composite operator insertions, this implies
$$
G_{n,B}=Z_\phi^{n/2}G_n.
$$
The bare <correlation function> is independent of the arbitrary <renormalization scale> $\mu$ when the bare parameters and external momenta are fixed. Define
$$
\widehat\beta_\lambda=\left.\mu\frac{d\lambda}{d\mu}\right|_B,\qquad \gamma_{m^2}=\left.\mu\frac{d\log m^2}{d\mu}\right|_B,\qquad \gamma_\phi=\left.\frac12\mu\frac{d\log Z_\phi}{d\mu}\right|_B.
$$
The definitions of the <running mass> and <wave-function renormalization> functions specify their signs and the factor of two. Differentiating $Z_\phi^{n/2}G_n$ using the <chain rule> yields the \b[Callan–Symanzik equation]
$$
\boxed{\left[\mu\partial_\mu+\widehat\beta_\lambda\partial_\lambda+m^2\gamma_{m^2}\partial_{m^2}+n\gamma_\phi\right]G_n=0.}
$$
Here $\widehat\beta_\lambda$ denotes the <renormalization-group beta function> itself, not the function multiplied by an additional $\lambda$. The <Callan-Symanzik equation> concerns dependence on the arbitrary reference scale, with the displayed external momenta kept fixed.
For the <minimal subtraction scheme> in <dimensional regularization>, write
$$
F(\lambda,\epsilon)=\lambda+\sum_{k\geq1}\frac{f_k(\lambda)}{\epsilon^k},\qquad H(\lambda,\epsilon)=1+\sum_{k\geq1}\frac{b_k(\lambda)}{\epsilon^k},
$$
so $\lambda_B=\mu^\epsilon F$ and $m_B^2=m^2H$. The $\mu^\epsilon$ factor gives the engineering contribution $-\epsilon\lambda$ to the <renormalization-group beta function>. Put $\widehat\beta_\lambda=-\epsilon\lambda+\beta_\lambda$. Scale independence of the bare coupling then gives
$$
0=\epsilon F+\widehat\beta_\lambda\partial_\lambda F
=\epsilon(F-\lambda F')+\beta_\lambda F'.
$$
The coefficient of $\epsilon^0$ is $f_1-\lambda f_1'+\beta_\lambda$. Its vanishing proves the first of the <simple-pole formulas for minimal-subtraction renormalization group functions>:
$$
\boxed{\beta_\lambda=\widehat\beta_\lambda+\epsilon\lambda=(\lambda\partial_\lambda-1)f_1(\lambda).}
$$
The coefficients of $\epsilon^{-k}$, $k\geq1$, also vanish and impose $(\lambda\partial_\lambda-1)f_{k+1}=\beta_\lambda f_k'$. These relations explain how the higher poles are consistent with a finite <renormalization-group beta function>.
For the mass, differentiate the second bare-parameter relation:
$$
0=\gamma_{m^2}H+\widehat\beta_\lambda H'.
$$
Its finite coefficient is $\gamma_{m^2}-\lambda b_1'$. Therefore
$$
\boxed{\gamma_{m^2}=\lambda\partial_\lambda b_1(\lambda).}
$$
The higher poles impose $\lambda b_{k+1}'=\gamma_{m^2}b_k+\beta_\lambda b_k'$. The formulas for the <running mass> and <renormalization-group beta function> thus follow directly from bare-parameter independence rather than from an identification with diagram coefficients by name.
At the specified <loop order>, $f_1=3\lambda^2/(16\pi^2)$ and $b_1=\lambda/(16\pi^2)$. There is no <wave-function renormalization> at this order. Hence
$$
\boxed{\beta_\lambda=\frac{3\lambda^2}{16\pi^2}+O(\lambda^3),\qquad \gamma_{m^2}=\frac{\lambda}{16\pi^2}+O(\lambda^2),\qquad \gamma_\phi=0+O(\lambda^2).}
$$
The regulated <renormalization-group beta function> is $\widehat\beta_\lambda=-\epsilon\lambda+3\lambda^2/(16\pi^2)+O(\lambda^3)$; after removing the regulator, $\widehat\beta_\lambda=\beta_\lambda$.
Now take $m^2=0$ and the four-dimensional limit. This massless surface is preserved since $\mu\,dm^2/d\mu=m^2\gamma_{m^2}$. Set $b=3/(16\pi^2)$, choose a reference scale $\mu_0$, and define the <running coupling> by
$$
\frac{d\lambda(\mu)}{d\log\mu}=b\lambda(\mu)^2,\qquad \lambda(\mu_0)=\lambda_0.
$$
Differentiating $1/\lambda$ gives $d(1/\lambda)/d\log\mu=-b$, so
$$
\boxed{\lambda(\mu)=\frac{\lambda_0}{1-b\lambda_0\log(\mu/\mu_0)}.}
$$
The zero-coupling solution is obtained by continuity. At one-loop accuracy $\gamma_\phi=0$, and the <chain rule> turns the massless <Callan-Symanzik equation> into
$$
\frac{d}{d\log\mu}G_n(p_1,\ldots,p_n;0,\lambda(\mu),\mu)=0.
$$
Thus the \b[one-loop solution along a running-coupling trajectory] is
$$
\boxed{G_n(p_1,\ldots,p_n;0,\lambda(\mu),\mu)=G_n(p_1,\ldots,p_n;0,\lambda_0,\mu_0).}
$$
For a prescribed endpoint coupling $\lambda$, substitute $\lambda_0=\lambda/[1+b\lambda\log(\mu/\mu_0)]$ on the right. The reference <correlation function> supplies arbitrary boundary data; the <Callan-Symanzik equation> determines their scale evolution, not their full momentum dependence. The <running coupling> formula is used on a branch on which its denominator is nonzero and the retained perturbative approximation is appropriate.
If one keeps $\epsilon\ne0$, the same characteristic calculation uses $d\lambda/d\log\mu=-\epsilon\lambda+b\lambda^2$. Writing $r=\mu/\mu_0$ gives $\lambda(\mu)=\lambda_0r^{-\epsilon}/[1-(b\lambda_0/\epsilon)(1-r^{-\epsilon})]$, which tends to the displayed four-dimensional result as $\epsilon\to0$.
With a nonzero <anomalous dimension>, the characteristic equation instead reads $dG_n/d\log\mu=-n\gamma_\phi(\lambda(\mu))G_n$. Integration proves the <characteristic solution of the massless Callan-Symanzik equation>:
$$
G_n(\boldsymbol p;0,\lambda(\mu),\mu)=\exp\left[-n\int_{\mu_0}^{\mu}\gamma_\phi(\lambda(\nu))\frac{d\nu}{\nu}\right]G_n(\boldsymbol p;0,\lambda_0,\mu_0).
$$
For the stipulated $\gamma_\phi=c\lambda^2$ and the same one-loop <renormalization-group beta function>, change variables using $d\log\nu=d\lambda/(b\lambda^2)$. The exponent's integral becomes $(c/b)(\lambda(\mu)-\lambda_0)$. Therefore the \b[modified explicit solution] is
$$
\boxed{G_n(\boldsymbol p;0,\lambda(\mu),\mu)=\exp\left[-\frac{nc}{b}\big(\lambda(\mu)-\lambda_0\big)\right]G_n(\boldsymbol p;0,\lambda_0,\mu_0).}
$$
Equivalently its multiplicative factor is $\exp[-nc\lambda_0^2\log r/(1-b\lambda_0\log r)]$. This solves the equation with the stated truncated <renormalization-group beta function> and <anomalous dimension>; it does not introduce additional higher-order coefficients that were not specified.
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