For a function $ f: \mathbb{R} \to \mathbb{R} $, write Graph($f$) $:= \{(x,f(x)):x\in \mathbb{R}\} $.
For $ f \in C^{\infty}(\mathbb{R}) $, define $ A(f) := \bigcup\limits_{n=0}^{\infty} \textrm{Graph}(f^{(n)}) $.
It is clear that $ g = f^{(n)} $ implies $ A(g) \subset A(f) $. The question is, is the converse also true? That is, does $ A(g) \subset A(f) $ imply $ g = f^{(n)} $ for some $ n \geq 0 $?
Let $g(x)=e^{- \frac 1 {x^2}}$ for $x \neq 0$, $g(0)=0$. Then $g \in C^{\infty} (\mathbb R)$ and all derivatives of g vanish at 0. Let $f(x)=g(x)$ for $x \leq 0$ and $f(x)=g'(x)$ for $x>0$. For any n and x $(x,f^{(n)} (x))=(x,g^{(n)} (x))$ or $(x,f^{(n)} (x))=(x,g^{(n+1)} (x))$ so $A(f) \subset A(g)$. However, there is no n such that $f=g^{(n)}$. (Sorry, I switched f and g!)