= Solution
Under the finite <dilatation>, $d^Dx'=e^{-D\alpha}d^Dx$, $\partial_\mu'\phi'(x')=e^{2\alpha}\partial_\mu\phi(x)$ and $\phi'(x')^n=e^{n\alpha}\phi(x)^n$. Keeping the couplings fixed gives
$$
S'=\int d^Dx\left[\frac12e^{(4-D)\alpha}(\partial\phi)^2-\frac{m^2}{2}e^{(2-D)\alpha}\phi^2-\sum_{n>2}\frac{\lambda_n}{n!}e^{(n-D)\alpha}\phi^n\right].
$$
Thus the full finite change is
$$
\boxed{\Delta S=\int d^Dx\left[\frac{e^{(4-D)\alpha}-1}{2}(\partial\phi)^2-\frac{m^2}{2}(e^{(2-D)\alpha}-1)\phi^2-\sum_{n>2}\frac{\lambda_n}{n!}(e^{(n-D)\alpha}-1)\phi^n\right].}
$$
For invariance for arbitrary fields and all constant $\alpha$, the fixed nonzero kinetic term requires $D=4$. Then the mass term requires $m=0$, and every nonzero coupling must have $n=4$. Hence
$$
\boxed{\mathcal L_{\rm scale}=\frac12\partial_\mu\phi\partial^\mu\phi-\frac{\lambda_4}{4!}\phi^4,\qquad D=4,\ m=0,}
$$
including the free choice $\lambda_4=0$. At $D=4$ the canonical <mass dimension> is $[\phi]=1$ and $[\lambda_n]=4-n$, so only the quartic interaction has a dimensionless coupling. A mass introduces a scale. This recovers the same result from <6/a/solution>.
The fixed field weight in the question matters: if it were replaced by the general canonical weight $(D-2)/2$, the kinetic term could be invariant in other dimensions, and a nonzero monomial interaction would require $n=2D/(D-2)$ for $D\ne2$. That is a different transformation from the one specified here. The result above is classical; quantum running generally breaks the quartic theory's <scale invariance>.
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