In the book from Coleman: The Aspects of Symmetry, p. 70; linear scale transformations or dilations are defined as $$ x \rightarrow e^\alpha x $$ with $\alpha$ being a real number. The fields change as $$ \phi(x) \rightarrow e^{\alpha d} \phi (e^{\alpha x}) $$ which yields an infinitesimal transformation $$ \delta \mathbf{\phi} = (d + x^\lambda \partial_\lambda ) \mathbf \phi $$where $d$ is a matrix. (It is clearer as $\delta \phi_i = (d_{ij} + x^\lambda \partial_\lambda \delta_{ij} ) \phi_j$, where $\delta_{ij}$ is the Kronecker delta.)
Now, there is a statement I have not been able to prove:
For a large class of theories (including all renormalizable field theories) these transformations are symmetries, if all non-dimensionless coupling constants (including the masses) are set equal to zero, and if $d$ is chosen to be a matrix that multiplies all Bose fields by one and all Fermi fields by $\frac{3}{2}$.
Based on this quote, I have tried using the lagrangian $\mathcal L = \partial_\mu \phi_i \partial^\mu \phi_i$ which is the Klein Gordon lagrangian with $m=0$. Computing $\mathcal L[\phi + \delta \phi] - L[\phi]$ gives the variation of the action which should be 0 up to a total derivative when $d_{ij} = 1 \cdot \delta_{ij}$. However I find terms with extra derivatives that do not cancel. I would like to know how to find the 1 and 3/2 for bosons and fermions. I guess this is a general result and there is no need to pick a specific lagrangian.