= Solution
{c}
Use the sign convention $E(\phi)=\delta S_0/\delta\phi$ from part <1/b>, and let
$$
S_\eta[\phi]=S_0[\phi]+\eta\int d^4x\,f(\phi)E(\phi).
$$
In the <path integral>, make the local <field redefinition> $\phi_{\rm old}=\chi-\eta f(\chi)$. The <functional Taylor expansion> of the <action> gives
$$
\begin{aligned}
S_0[\chi-\eta f(\chi)]
&=S_0[\chi]-\eta\int d^4x\,f(\chi)\frac{\delta S_0}{\delta\chi(x)}+O(\eta^2),\\
\eta\int d^4x\,f(\phi_{\rm old})E(\phi_{\rm old})
&=\eta\int d^4x\,f(\chi)E(\chi)+O(\eta^2).
\end{aligned}
$$
The two first-order terms cancel. Neglecting the <functional Jacobian> as permitted,
$$
\boxed{Z_\eta=\mathcal N\int\mathcal D\chi\,\exp\{iS_0[\chi]+O(\eta^2)\}.}
$$
Thus $fE$ is a <redundant operator> at first order, by the <equivalence theorem for field redefinitions>. The same argument works for a local derivative-dependent $f$, which is needed to remove $(\Box\phi)^2$. Insertions and sources must be transformed too: arbitrary untransformed <off shell> <correlation functions> need not be equal.
\b[Sign qualification.] If instead $E_+=\Box\phi+m^2\phi+\lambda\phi^3/6$ is used, then $E_+=-\delta S_0/\delta\phi$. An added $+\eta fE_+$ is removed by $\phi_{\rm old}=\chi+\eta f(\chi)$. The minus shift requested in the source therefore fixes which sign is meant by its otherwise unspecified $E$.
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