This is problem 9 from page 264 of Pugh's Real Mathematical Analysis. If $f:\mathbb{R}\mapsto \mathbb{R}$ continuous and the sequence of functions $f_n(x)=f(nx)$ is equicontinuous then what can be said about $f$?
My gut tells me that $f$ must be a constant function, and I've tried to prove it by showing that $f'$ must be zero everywhere, but I haven't been successful. Could someone help me prove this, or tell me what is actually going on here?
Your gut is right: $f$ must be constant. Fix $x_0,y_0\in\mathbb R$ and take any $\varepsilon>0$. By equicontinuity, there must exist some $\delta>0$ such that \begin{align*} \left.\begin{array}{l}x,y\in\mathbb R\\|x-y|<\delta\end{array}\right\}\quad\text{imply that}\quad|f(nx)-f(ny)|<\varepsilon\text{ for all $n\in\mathbb N$.} \end{align*} Now take $n\in\mathbb N$ so large that $|x_0-y_0|<n\delta$. Since $$\left|\frac{x_0}{n}-\frac{y_0}{n}\right|<\delta,$$ it follows that $$|f(x_0)-f(y_0)|=\left|f\left(n\frac{x_0}{n}\right)-f\left(n\frac{y_0}{n}\right)\right|<\varepsilon.$$ Since $\varepsilon>0$ can be arbitrarily small, one can conclude that $f(x_0)=f(y_0)$.