Consider Euclidean Klein Gordon quantum field theory on the toroidal spacetime $X\simeq S^1\times \cdots\times S^1$, with action
$$S(\varphi) = \int_X \varphi(\Delta+m^2)\varphi$$
and scalar field $\varphi\in C^\infty_X$, the space of smooth functions on $X$. Here, $\Delta$ is the Laplace operator corresponding to the standard flat metric on the torus. The claim is that this theory is invariant under the Wilsonian renormalization semigroup.
To fomulate this statement precisely, let $C^\infty_{(a,b)}$ denote the linear span of smooth functions with eigenvalue in $(a,b)$ under application of $\Delta$, and decompose the space of smooth functions according to the eigenvalues of the Laplace operator:
$$C^\infty_{[0,\Lambda)} \simeq C^\infty_{[0, \Lambda')}\oplus C^\infty_{[\Lambda',\Lambda)}.$$
Using the formula for the action of the Wilsonian renormalization semigroup $S[\Lambda]\to S[\Lambda']$ on the space of quantum field theories as given in Renormalization and Effective Field Theory, we have, for $\varphi\in C^\infty_{[0, \Lambda')}$,
$$S[\Lambda'](\varphi)=\frac{\hbar}{i}\log \bigg(\int_{\varphi^\perp \in C^\infty_{[\Lambda',\Lambda)}}d\mu^\perp \exp(iS[\Lambda](\varphi+\varphi^\perp)/\hbar)\bigg),$$
where we have factored the low-energy Feynman measure out from the higher-energy Feynman measure via $d\mu_\Lambda\equiv d\mu_{\Lambda'}\wedge d\mu^\perp$, where the Feynman measure $d\mu_\lambda$ at a general energy scale $\lambda$ is defined by the condition
$$\int_{\varphi\in C^\infty_{[0,\lambda)}}d\mu_\lambda \exp(iS[\lambda](\varphi)/\hbar)\equiv 1. $$
First note that the operator $(\Delta+m^2)$ lies in the algebra generated by the operators $1,\Delta$, and therefore trivially respects the eigenspace decomposition of $\Delta$. Therefore, the modes of the scalar field are uncoupled, and the action factorizes as
$$S[\Lambda](\varphi+\varphi^\perp)=S[\Lambda](\varphi)+S[\Lambda](\varphi^\perp),$$
and therefore the functional integral also factorizes:
$$S[\Lambda'](\varphi)=S[\Lambda](\varphi)+\frac{\hbar}{i}\log \bigg(\int_{\varphi^\perp \in C^\infty_{[\Lambda',\Lambda)}}d\mu^\perp \exp(iS[\Lambda](\varphi^\perp)/\hbar)\bigg),$$
and therefore the action gains at most a constant offset, which nonetheless leaves the overall theory invariant. Therefore massive Klein Gordon theory, and, as it seems, any other free theory which does not couple low- and high-energy degrees of freedom, must be conformally invariant, that is, invariant under Wilson's renormalization semigroup. This statement, however, seems to contradict the literature, so I guess I am looking for the source of my misunderstanding.