Suppose $f\colon M\to\mathbb{R}$ is a smooth 0-form on our manifold $M$, and $X$ is a vector field with an induced flow $\varphi_t\colon M\to M$.
I am wanting to show that $\mathcal{L}_Xf = f$ implies that $\varphi_t^* f = e^t f$ but I am having trouble making any progress.
What I know is that we define the Lie derivative as $$\mathcal{L}_Xf := \lim_{t\to 0}\frac{\varphi_t^*f - f}{t},$$ but the problem is that I do not know exactly how to manipulate it into the desired form. If $$f(p)= \lim_{t\to 0}\frac{f(\varphi_t(p)) - f(p)}{t},$$ I don't exactly have any leads. Perhaps I am looking at it from the wrong angle and I should not use the limit definition?
I would be particularly happy with any hints, as I do not want to spoil the exercise too much. Thanks!
Let $p \in M$ be fixed. Let $I \subset \mathbb{R}$ be the domain of $t \mapsto \phi_t(p)$, so that $I$ is an open interval containing $0$, and define a function $g_p : I \to \mathbb{R}$ by $g_p(t) := \phi_t^\ast f(p) = f(\phi_t(p))$. By the chain rule (or by the definition of smooth function on a manifold, depending on what reference you're using) the function $g_p$ is differentiable on all of $I$. What, then, is the relationship between the derivative $g_p^\prime : I \to \mathbb{R}$ of $g_p$ and $(\mathcal{L}_X f)(p)$, and what does it tell you about $g_p$? Note that this may require a tiny bit of care for $t \neq 0$.