= Solution
Set $B(x)=\bar\psi(x)\psi(x)$, a Grassmann-even <Dirac scalar bilinear>. Completing the square gives
$$
-\frac12\sigma^2+g\sigma B
=-\frac12(\sigma-gB)^2+\frac{g^2}{2}B^2.
$$
Introduce external Grassmann <sources> $\eta,\bar\eta$ for the <Dirac field>, but no source for $\sigma$. The <generating functional> is
$$
Z[\eta,\bar\eta]=\int\mathcal D\bar\psi\,\mathcal D\psi\,\mathcal D\sigma\,
\exp\left\{i\int d^4x\left[\bar\psi(i\not\partial-m)\psi
-\frac12(\sigma-gB)^2+\frac{g^2}{2}B^2
+\bar\eta\psi+\bar\psi\eta\right]\right\}.
$$
At a finite regulator the change of variable $\sigma'=\sigma-gB$ has unit Jacobian: it translates a bosonic integration variable by a Grassmann-even expression independent of that variable. Its oscillatory <Gaussian integral> can be defined with a convergence prescription and then continued. It is a constant $\mathcal N$ independent of $\psi,\eta,\bar\eta$. Thus
$$
Z[\eta,\bar\eta]=\mathcal N\int\mathcal D\bar\psi\,\mathcal D\psi\,
\exp\left\{i\int d^4x\left[\bar\psi(i\not\partial-m)\psi
+\frac{g^2}{2}(\bar\psi\psi)^2+\bar\eta\psi+\bar\psi\eta\right]\right\}.
$$
The factor $\mathcal N$ cancels in $Z[\eta,\bar\eta]/Z[0,0]$. Every normalized <Green function> with only external <fermion> lines is a Grassmann source derivative of this same normalized <generating functional>, so
$$
\boxed{G_{\mathrm{auxiliary}}^{(n)}=G_{\mathrm{four\text{-}fermion}}^{(n)}.}
$$
This is an exact <Hubbard–Stratonovich transformation> in reverse, rather than just an approximation based on <tree-level Feynman diagrams>. At the diagrammatic level the contact <auxiliary-field propagator> $-i\delta^{(4)}(x-y)$ collapses each exchanged line into the same local <four-fermion interaction>. The equality uses a common regulator and does not assert perturbative <renormalizability> of that four-dimensional <four-fermion interaction>.
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