Question: So we all know that $\frac d{dx}e^x=e^x$ and that the $n$th derivative of $e^x$ is still $e^x$, but upon entering fractional calculus, this is ruined. Let $D^\alpha$ be the $\alpha$th derivative with respect to $x$. Then, [as we can see](https://math.stackexchange.com/questions/1322857/taking-the-half-derivative-of-ex), when $\alpha\in[0,1)$,

$$D^\alpha e^x=-\frac1{\Gamma(1-\alpha)}e^x\gamma(\alpha,x)\ne e^x$$

where we use the lower incomplete gamma function.

Which raises the interesting question:

> 
> What are the solutions to the following fractional differential equation? $$D^\alpha f(x)=f(x)$$
> 
> 
> 

where we have

$$D^\alpha f(x)=\frac1{\Gamma(n-\alpha)}\int\_0^x\frac{f^{(n)}(t)}{(x-t)^{\alpha+1-n}}\ dt$$

with $n=\lfloor\alpha+1\rfloor$.

---

$f$ may be a function of $\alpha$.

Answer: 
This is a (heuristic) answer for $\alpha=1/k$ for $k\in \mathbb{N}$. And assuming $Df$ exists. Note that
I believe (see edit in the end) that we could extend this answer to any $\alpha$.

Let $\alpha = 1/2$. Then, by definition of the fractional derivative, we have
$$D^\alpha D^\alpha f = Df$$
Substituting on the LHS what the differential equation implies we obtain
$$D^\alpha f = D f$$

More general,
$$D^{(k-1)\alpha}f=D f \quad (\*)$$

I can only think of the function $f\equiv 0$ which satisfies this and the given differential equation.

**Edit:** We could continue with equation $(\*)$ by plugging in repetedly the diff eq on the LHS to obtain
$$ D^{(k-2)\alpha}f=D f$$
$$ D^{(k-3)\alpha}f=D f$$
$$\ldots$$
 $$ f=D f$$
Since we know two solutions for the very last equation and since one of them does not satisfy the fract diff eq given in the question, we conclude that indeed $f\equiv 0$ is the only soluton. 

I believe that this is also true for any $\alpha$ if we assume that $f$ has a derivative up to a certain order: Start with rational $\alpha=p/q$ then start the argument with $ (D^\alpha)^q f$.
IDK if $\alpha<1$

For special cases of $\alpha >1$,

$f^\alpha=f$ gives: 

$f=\sum\_{k\leq\alpha} C\_ke^{e^{\frac{{2ki\pi}}{\alpha}}x}+\sum\_{n\leq\alpha} C\_ne^{e^{\frac{{2i\pi}}{n}}x}$
where $k,n\in\mathbb{N}^\*$