The simplest example of this is $e^x$ which we could say has period 1 (it is its own derivative). $e^{-x}$ would have period 2.
Using similar constructions, I can get a function that has a derivative of period $n$ by doing $e^{x\cdot (1)^{1/n}}$
Is this the only way to get periodic derivatives?
Note: I am treating sin and cos as special cases of this when $n=4$
Is there a proof to this effect?
Assume $f^{(n)}(x)=f(x)$ for all $x$. Let $g(x) = \sum_{k=0}^{n-1}\xi^{-k} f^{(k)}(x)$, where $\xi^n=1$. Then $g'(x)=\sum_{k=0}^{n-1}\xi^{-k} f^{(k+1)}(x) = \xi\sum_{k=1}^{n}\xi^{-k} f^{(k)}(x)=\xi g(x)$. We already know all solutions of $y'=c y$ and conclude that $g(x)=a e^{\xi x}$.
If we additionally assume that $\xi$ is primitive, we find for $0\le m<n$ that $$ \sum_{k=0}^{n-1}\xi^{-mk} f^{(k)}(x)= a_m e^{\xi^m x}$$ Adding all equations leads to $$n f(x) = \sum_{m=0}^{n-1} a_m e^{\xi^m x}.$$