In some field theory textbooks, such as the CFT Yellow Book (P40), the authors claim that a theory has a certain symmetry, which means that the action of the theory does not change under the symmetry transformation. For example, given a theory $\mathcal{L}(\phi,\partial\phi)$ living in a $d$-dimensional flat space, if the following coordinates mapping and the associated field transformation $$ x\to x'=f_w(x),\quad\phi(x)\to\phi'(x')=F_w[\phi(x)]\tag{1} $$ form a symmetry transformation, where the subscript $w$ parameterizes the symmetry, then we have
\begin{align} S_{V'}[\phi']\equiv\int_{V'}d^dx'\mathcal{L}(\phi',\partial'\phi')=S_{V}[\phi]\equiv\int_{V}d^dx\mathcal{L}(\phi,\partial\phi),\tag{2} \end{align}
where $V'$ is the image of $V\subseteq\mathbb{R}^d$ under the mapping $f_w$. We can obtain the following reasonable conclusion from (2):
Conclusion 1. if the field configuration $\phi^{\star}$ is on-shell, i.e., $\phi^\star$ satisfies the equation of motion (EoM), then $F_w[\phi^\star]$ is also on-shell.
proof . since $S_V[\phi^\star+\delta\phi]$ (where $\delta\phi|_{\partial V}=0$) gets its minimum at $\delta\phi=0$, then using eq.(2) we know that $S_{V'}[F_w[\phi^\star+\delta\phi]]=S_{V'}[F_w[\phi^\star]+\delta\phi']$ (where we have $\delta\phi'|_{\partial V'}=0$) gets its minimum at $\delta\phi'=0$. Therefore $\phi'$ is on-shell $\square$.
However, in some other textbooks, such as in Peskin's QFT book (P17) and Banados et al's review of Noether's theorem (P17), I found that the definition of symmetry is more relaxed. The action before and after the symmetry transformation (1), in their statements, can be up to a surface term, that is, we have $$ S_{V'}[\phi']=S_{V}[\phi]+\int_{\partial V}dS_{\mu}K^\mu(\phi,\partial\phi). \tag{3} $$ However, once $K^{\mu}$ is allowed to be a function of $\partial\phi$, I cannot reproduce conclusion 1. The reason is that when I imitated the proof above, I found that $$ \int_{\partial V}dS_{\mu}K^\mu(\phi^\star+\delta\phi,\partial\phi^\star+\partial\delta\phi) $$ is not $\delta\phi$-independent. Because we only imposed the condition of $\delta\phi=0$ on the boundary $\partial V$, but not the condition $\delta\partial\phi|_{\partial V}=0$. Therefore in this case I can't prove that $S_{V'}[F_w[\phi^\star]+\delta\phi']$ gets its minimum at $\delta\phi'=0$.
Here is my question: Can we reproduce conclusion 1 in the relaxed symmetry definition (3) without adding more conditions like $\delta\partial\phi|_{\partial V}=0$?
Note added: I found a very similar Post in Phys.SE "Invariance of action $\Rightarrow$ covariance of field equations?" In that question, @Qmechanic called this relaxed symmetry definition (3) quasi-symmetry and gave a positive answer. However, as he said, he dropped the boundary contribution in the derivation, thus making the proof less rigorous.
