Suppose we have a function $F:M\rightarrow\mathbb{R}^l$ from a smooth $m$-manifold $M$ to $\mathbb{R}^l$, where $l\le m$. If we denote the components of $F$ by $F^{\nu}$, we can define a system of $l$ equations:
$$F^{\nu}(x)=0.\tag{1}$$ for $\nu=1,\ldots,l$.
Let $G$ be a (connected local) Lie group of transformations on $M$, suppose we want to find the condition for the solution set $(1)$ to be invariant under $G$. Note that $F$ itself need not be $G$-invariant. I would naively expect that the solution set to $(1)$ would be invariant iff $w(F^{\nu})=0$ for every infinitesimal generator $w$ of $G$, since (representing the exponential map as $e^{\epsilon w}$) $$F^{\nu}(e^{\epsilon w}x)=F^{\nu}(x)+w(F^{\nu})(x)\epsilon+o(\epsilon),\tag{2}$$ and so $$\left.\frac{d}{d\epsilon}F^{\nu}(e^{\epsilon w}x)\right\rvert_{\epsilon=0}=w(F^{\nu})(x).\tag{3}$$ The left hand side will be zero iff the right hand side is.
However, it turns out (Theorem 2.8 in Olver's "Applications of Lie Groups to Differential Equations") that in addition to $(3)$ you also need the differential $DF$ to have maximum rank (on the solution set $(1)$). Could someone please tell me what I'm missing that makes (1-3) invalid?
The condition $$\left.\frac{d}{d\epsilon}F^{\nu}(e^{\epsilon w}x)\right\rvert_{\epsilon=0}=0$$ at the point $x$ alone isn't enough to guarantee that $F^\nu(e^{\epsilon w}x)=0$ for $\epsilon > 0,$ so as written your argument only shows that this condition is necessary, not sufficient. It shouldn't be too surprising that we need a full-rank assumption to make both directions work - this is exactly the assumption we need to invoke something like the implicit function theorem in order to "straighten out" $F$. In particular, it implies that any $w$ with $w(F^\nu)=0$ is tangent to some level curve of $F^\nu,$ which is probably part of your mental picture.
You can try thinking about some examples: the most obvious would be something like $F(x) = x^2,$ where the solution set is $\{ 0 \}.$ Any group of transformations that moves this point (e.g. the translations) is then a counterexample, since $DF|_0 = 0.$