= Solution
Under the finite <dilation>, the <volume element> transforms as $d^4x'=e^{4\alpha}d^4x$, whereas $\partial'_\mu\phi'(x')=e^{-2\alpha}\partial_\mu\phi(x)$. Thus the <kinetic term> and <quartic interaction> both acquire the factor $e^{-4\alpha}$. Their factors cancel the <Jacobian determinant>, proving
$$
\boxed{S[\phi']=S[\phi]\quad(m=0).}
$$
A <mass term> would instead acquire a net factor $e^{2\alpha}$ in the <action> and break this <scale invariance>.
For the <Noether current>, use the variation at the same coordinate, rather than comparing values at different coordinates:
$$
\delta_0\phi(x)=\phi'(x)-\phi(x)
=-\alpha\bigl(\phi+x^\nu\partial_\nu\phi\bigr).
$$
At $m=0$, the corresponding <Lagrangian density> variation is $\delta_0\mathcal L=-\alpha(4+x\cdot\partial)\mathcal L=-\alpha\partial_\mu(x^\mu\mathcal L)$. Thus $J_0^\mu=-x^\mu\mathcal L$, and the stated <Noether current> convention gives
$$
\boxed{j^\mu=-x_\nu T^{\mu\nu}-\phi\partial^\mu\phi.}
$$
The frequently used <dilatation current> $D^\mu=x_\nu T^{\mu\nu}+\phi\partial^\mu\phi$ is $-j^\mu$; this is solely an overall parameter convention. In the convention $D^\mu=x_\nu T^{\mu\nu}-V^\mu$, the <virial current> is $V^\mu=-\phi\partial^\mu\phi$.
Using <1/b/solution> and the <trace> computed in <1/c/solution> gives the useful identity
$$
\partial_\mu D^\mu
=T^\mu{}_{\mu}+x^\nu F\partial_\nu\phi+(\partial\phi)^2+\phi\Box\phi
=m^2\phi^2+\bigl(\phi+x^\nu\partial_\nu\phi\bigr)F.
$$
It follows that $\boxed{\partial_\mu j^\mu=0}$ in the massless theory <on shell>. Equivalently, $x_\nu\Omega^{\mu\nu}$ is a <conserved current>, since its <divergence> is the <trace> of the improved <stress-energy tensor> <on shell>. These are classical <scale invariance> statements; the nonzero <beta function> in <4/d/solution> explains why the interacting quantum theory need not retain them.
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