= Solution
For a connected amputated <Feynman diagram>, write $L$ for the number of independent <loop momenta>. Each <loop momentum> integration scales as $\lambda^4$, each internal <scalar propagator> as $\lambda^{-2}$, and each internal <Dirac propagator> as $\lambda^{-1}$ when both <four-momenta> and <mass> are scaled. The <Yukawa coupling> has <mass dimension> zero. Thus
$$
d_S=4L-n_f-2n_s.
$$
Counting the two <fermion> ends and one scalar end at each <interaction vertex>, and using connectedness, gives
$$
2n_f=2n_v,\qquad 2n_s+n=n_v,\qquad
L=n_f+n_s-n_v+1.
$$
Consequently $n_f=n_v$, $L=n_s+1$, and
$$
\boxed{d_S=4+2n_s-n_v=4-n.}
$$
\b[The requested exponent requires amputation and removal of the overall momentum delta function.] An ordinary unamputated connected <Green function> has an additional $n$ external <scalar propagators>, giving <mass dimension> $4-3n$ after stripping that delta function, or $-3n$ if it is retained. For a concrete check, the free two-point <scalar propagator> scales as $\lambda^{-2}$, not $\lambda^2$. The exponent $4-n$ is the engineering result for the amputated object. A fixed <renormalization scale> introduces logarithms; the displayed homogeneity is a naive <power counting> statement rather than an exact quantum scaling law at fixed subtraction scale.
More generally, with $E_s$ external scalar ends and $E_f$ external <fermion> ends, the same counting gives the <superficial degree of divergence>
$$
\boxed{D=4-E_s-\frac32E_f.}
$$
This is an upper bound on the overall <ultraviolet divergence>. Numerator cancellations and <Lorentz invariance> may lower it; an apparently convergent whole <Feynman diagram> can still have <ultraviolet subdivergences>. Local subtraction of those <ultraviolet subdivergences> is required first.
For pure scalar ends, $E_s=0,1,2,3,4$ can require <counterterms>. <Lorentz invariance> reduces the two-point terms to a <mass counterterm> and <wave-function renormalization>; derivatives of a single scalar field are total derivatives, and a one-derivative cubic scalar term cannot be a <Lorentz scalar>. With two external <fermions>, $E_s=0$ requires their <kinetic term> and <mass counterterm>, while $E_s=1$ requires <Yukawa coupling> renormalization. Four external <fermions> give $D=-2$, so no independent four-<fermion> <counterterm> is required.
A complete <counterterm Lagrangian> consistent with the species symmetry can therefore be written
$$
\boxed{\begin{aligned}
\mathcal L_{\rm ct}={}&\frac{\delta Z_\phi}{2}(\partial\phi)^2
-\frac{\delta m_s^2}{2}\phi^2-\delta t\,\phi
-\frac{\delta\kappa}{3!}\phi^3-\frac{\delta\lambda_4}{4!}\phi^4-\delta\mathcal E\\
&+\sum_{i=1}^2\bar\psi_i(\delta Z_\psi i\not\partial-\delta m_f)\psi_i
-\delta g\,\phi\sum_{i=1}^2\bar\psi_i\psi_i.
\end{aligned}}
$$
The <vacuum energy density> term matters for an unnormalized <vacuum functional> and cancels from normalized <Green functions>. The <tadpole counterterm> can impose $\langle\phi\rangle=0$. There is no $\phi\mapsto-\phi$ symmetry of the massive <Yukawa interaction>, so linear, cubic and quartic scalar terms cannot be excluded. For instance, in a constant background the divergent <fermion loop> contribution contains $(m+g\phi)^4$, exhibiting all these powers. This is <counterterm closure of a massive Yukawa theory>.
The two species have identical couplings, so their common unitary flavour rotations preserve the <Lagrangian density> and their <counterterms> can be chosen species independent. Equality of the scalar and <fermion> masses, however, is not protected: $\delta m_s^2$ and $\delta m_f$ are independent. Including the displayed finite set of operators yields a <renormalizable quantum field theory>. Restricting the bare <Lagrangian density> to the terms printed in the question is not closed under <renormalization>.
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