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The finite passive dilatation is $x^{\prime }=e^{-\alpha }x$, ${\varphi }^{\prime }(x^{\prime })=e^{\alpha }\varphi (x)$. To compare at the same argument, invert the coordinate change:Consequently the fixed-argument infinitesimal variation isThis also follows from $\delta \varphi =\alpha \varphi$ and $\delta x^{\mu }=-\alpha x^{\mu }$. The plus sign on the derivative term is a consequence of this specified contraction of coordinates.

Under the finite dilatation, $d^D{x}^{\prime }=e^{-D\alpha }{d}^{D}x$, ${\partial }_{\mu }^{\prime }{\varphi }^{\prime }(x^{\prime })=e^{2\alpha }{\partial }_{\mu }\varphi (x)$ and ${\varphi }^{\prime }(x^{\prime }{)}^n=e^{n\alpha }\varphi (x{)}^n$. Keeping the couplings fixed givesThus the full finite change isFor 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 $n=4$. Henceincluding the free choice ${\lambda }_4=0$. At $D=4$ the canonical mass dimension is $[\varphi ]=1$ 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 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 $D\ne 2$. 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.