Derivative of a function is same the function Lets extend to nth order derivative  functions and not merely second order derivative functions.
The 4th order(say) derivative of the following functions is same the function itself
$$
\begin{aligned}
f(x)&=e^{x}\\
f(x)&=0\\
f(x)&=\sin x\\
f(x)&=\cos x
\end{aligned}
$$
I was curious to know whether there are there any more such unique functions ?
 A: This type of problem is best approached via the characteristic equation. In this case, we have $y^{(4)} = y$, which yields characteristic equation $r^4 - 1 = 0$. The roots of this equation over $\mathbb{C}$ are $r = \pm 1, \pm i$, so the general solution to the differential equation is
$$y = A_1e^{x} + A_2e^{-x} + A_3e^{ix} + A_4e^{-ix}, \quad A_i \in \mathbb{C}.$$
Using $\cos(x) = (e^{ix} + e^{-ix})/2$ and $\sin(x) = (e^{ix} - e^{-ix})/(2i)$ explains why your solutions arise for certain choices of the numbers $A_i$, and allows us to conclude that your list is almost complete. (You just forgot $e^{-x}$.) In other words, $e^{x}, e^{-x}, \cos(x), \sin(x)$ form a $\textit{basis}$ for the solution space.
A: Suppose you were looking for solutions to $\dfrac{d^ny}{dx^n}=y$.  Obvious solutions might be:


*

*$y_0 = 1 +\dfrac{x^{n}}{n!}  +\dfrac{x^{2n}}{(2n)!}+\dfrac{x^{3n}}{(3n)!}+\cdots$

*$y_1 = x +\dfrac{x^{n+1}}{(n+1)!}  +\dfrac{x^{2n+1}}{(2n+1)!}+\dfrac{x^{3n+1}}{(3n+1)!}+\cdots$

*$y_2 = \dfrac{x^{2}}{2!}  +\dfrac{x^{n+2}}{(n+2)!}  +\dfrac{x^{2n+2}}{(2n+2)!}+\dfrac{x^{3n+2}}{(3n+2)!}+\cdots$

*$\cdots$

*$y_{n-1} = \dfrac{x^{n-1}}{(n-1)!}  +\dfrac{x^{2n-1}}{(2n-1)!}+\dfrac{x^{3n-1}}{(3n-1)!}+\dfrac{x^{4n-1}}{(4n-1)!}+\cdots$


and in fact these $\displaystyle y_j= \sum_{k=0}^{\infty} \dfrac{x^{kn+j}}{(kn+j)!}$ for $0 \le j \lt n$ and linear combinations of these $y_j$s are the only solutions to $\dfrac{d^ny}{dx^n}=y$ (including the constant $0$). 
For example with $n=4$, you have solutions such as 


*

*$e^x=y_0+y_1+y_2+y_3$

*$e^{-x}=y_0-y_1+y_2-y_3$

*$\sin(x)=y_1-y_3$

*$\cos(x)=y_0-y_2$

*$\sinh(x)=y_1+y_3$

*$\cosh(x)=y_0+y_2$


but you could also have something like $7y_1-2y_2$ as a solution
