= Solution
Let $b,c>1$, rescale $q'_\parallel=bq_\parallel$, $\mathbf q'_\perp=c\mathbf q_\perp$, and set $B=bc^{d-1}$. Use the Fourier-amplitude convention
$$
\widetilde m'(q')=\widetilde m(q)/z.
$$
The momentum measure contributes $B^{-1}$, and substitution into the quadratic <statistical Hamiltonian> gives
$$
\boxed{K'=\frac{z^2}{Bb^2}K,\quad L'=\frac{z^2}{Bc^4}L,\quad
t'=\frac{z^2}{B}t,\quad h'=zh.}
$$
In the uniform source the momentum delta function cancels the <Jacobian determinant> of the measure, explaining why $h'$ does not have an additional $B^{-1}$. In a full <momentum-shell renormalization group> transformation integrate out the fast modes first. Their uncoupled <Gaussian integrals> contribute an additive <free-energy density>; the displayed transformations describe the slow-mode <statistical Hamiltonian> after coordinate and field rescaling.
The field convention must be stated. In real space, $x'_\parallel=x_\parallel/b$ and $\mathbf x'_\perp=\mathbf x_\perp/c$, and Fourier inversion gives $m'(x')=Bm(x)/z$. Thus if instead the symbol $m'=m/z_x$ refers to real-space amplitudes, $z=Bz_x$. Then the same transformation reads $K'=Bz_x^2K/b^2$, $L'=Bz_x^2L/c^4$, $t'=Bz_x^2t$, $h'=Bz_xh$. These are equivalent normalizations, not different <critical exponents>.
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