= Solution
For a translation-invariant potential, the on-shell <energy-momentum tensor> obeys $\partial_\mu\theta^{\mu\nu}=0$. The <divergence> of the <dilatation current> is therefore
$$
\partial_\mu j^\mu=(\partial\phi)^2+\phi\Box\phi+\theta^\mu{}_{\mu},\qquad\theta^\mu{}_{\mu}=\left(1-\frac D2\right)(\partial\phi)^2+DV(\phi).
$$
Use the <equation of motion> $\Box\phi=-V'(\phi)$ to obtain the useful general identity
$$
\boxed{\partial_\mu j^\mu=\frac{4-D}{2}(\partial\phi)^2+\left(\frac D2-1\right)m^2\phi^2+\sum_{n>2}(D-n)\frac{\lambda_n}{n!}\phi^n\quad\text{on shell}.}
$$
For $D=4$, $m=0$ and only $\lambda_4$ nonzero, every term vanishes. More explicitly, $\phi\Box\phi=-\lambda_4\phi^4/6$ while $\theta^\mu{}_{\mu}=-(\partial\phi)^2+\lambda_4\phi^4/6$; adding the first kinetic term cancels both remaining contributions. Thus $\partial_\mu j^\mu=0$, and $\int d^3x\,j^0$ is conserved when the boundary flux vanishes. Outside the parameter conditions of <6/d/solution>, the boxed formula displays the classical sources of scale breaking; the same current need not be conserved.
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