Remark: I think my answer should be ignored. (Apparently I did not understand the problem properly. Probably $S$ should be a linear functional on $C^\infty(M)$ (time-dependent?) etc. I still believe that the question ultimately boils down to something elementary once it is formulated in the right way. I just don't understand how the objects are defined.)
I think the question would be easier to answer for mathematicians if formulated in standard math language. I am not sure I am able to translate it, but let me try: You have some configuration space (probably a manifold, maybe infinite-dimensional) $M$, and a one-parameter group ${g_t}$ of $M$ (probably a diffeomorphism) and I guess that you assume that this is contained in some nice Lie group $G$ so that $g_t = \exp(tX)$ with $X \in \mathfrak g$, the Lie algebra of $G$. Now you have a (smooth?) function S on M and you want it to be invariant under ${g_t}$. This just means $XS = 0$. Now what does OS mean? Maybe you could explain this. It seems to me that you want to deduce something about $X$ and the derivative of $S$, but I am not quite sure I understand your notation there. I believe if you reformulate your question along these lines, more people can help.
I think the question would be easier to answer for mathematicians if formulated in standard math language. I am not sure I am able to translate it, but let me try: You have some configuration space (probably a manifold, maybe infinite-dimensional) $M$, and a one-parameter group ${g_t}$ of $M$ (probably a diffeomorphism) and I guess that you assume that this is contained in some nice Lie group $G$ so that $g_t = \exp(tX)$ with $X \in \mathfrak g$, the Lie algebra of $G$. Now you have a (smooth?) function S on M and you want it to be invariant under ${g_t}$. This just means $XS = 0$. Now what does OS mean? Maybe you could explain this. It seems to me that you want to deduce something about $X$ and the derivative of $S$, but I am not quite sure I understand your notation there. I believe if you reformulate your question along these lines, more people can help.