Let $f: \mathbb R \to \mathbb R$ be $C^\infty$. Suppose $f$ is not constant on any interval containing $x_0$. Are the following two statements equivalent:
(i) All derivatives of $f$ at $x_0$ vanish.
(ii) The rate of change of $f$ at $x_0$ is smaller than any polynomial function. More formally, if $g(h) := f(x_0 + h) - f(x)$, for any polynomial function $P(h)$, for any interval $I$ containing $x_0$, there exists $x_1 \in I$ such that for all $h < |x_1 - x_0|, 0 < |g(h)| < |P(h)|$.
Background: I am exploring flat functions and conjecture that the "reason" their derivatives vanish despite the function being non-constant is that their rate of change is too small to be captured by any polynomial. I am drawing on asymptotic concepts such as $O, o$, etc.
Let me first restate your condition (ii) to correct various minor errors which I don't think you intended:
By Taylor's theorem, the first $n$ derivatives of $f$ vanish at $x_0$ iff $g(h)/h^n\to 0$ as $h\to 0$. (Alternatively, instead of invoking Taylor's theorem, you can just prove this directly by induction on $n$ by applying L'Hopital's rule to $\lim_{h\to 0}g(h)/h^n$, which tells you that the limit is equal to $f^{(n)}(x_0)/n!$ as long as the first $n-1$ derivatives vanish.) So it suffices to show that this condition (ii) is equivalent to $g(h)/h^n\to 0$ for all $n$. In one direction, if you know that for all $\epsilon>0$, $|g(h)|<|\epsilon h^n|$ for $h$ sufficiently close to $0$, this means $g(h)/h^n\to 0$ as $h\to 0$. Conversely, if $P$ is a nonzero polynomial of degree less than $n$, then $h^n/P(h)\to 0$ as $h\to 0$ (for instance, by L'Hopital's rule). Consequently, if $g(h)/h^n\to 0$, then $g(h)/P(h)\to 0$ as well, which in particular implies $|g(h)|<|P(h)|$ for $h$ sufficiently close to $0$.