Solutions to this fractional differential equation 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, 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 $\alpha$.
 A: The first part is not a strict answer to the question, but not far. The full answer is added in second part.
Consider the series expansion :
$$e^x=\sum_{k=0}^\infty \frac{x^k}{k!} = \sum_{k=0}^\infty \frac{x^k}{\Gamma(k+1)} \qquad |x|<1$$
Compare to the Mittag-Leffler function :
$$E_\alpha(x)=\sum_{k=0}^\infty \frac{x^k}{\Gamma(\alpha k+1)}$$
http://mathworld.wolfram.com/Mittag-LefflerFunction.html
$$\text{Or }\qquad E_\alpha(x^\alpha)=\sum_{k=0}^\infty \frac{x^{\alpha k}}{\Gamma(\alpha k+1)}$$
This function matches the exponential function in particular case $\alpha=1$.
It is of interest to see what is the fractional derivative of $\left(E_\alpha(x^\alpha)-1\right)$. We will see latter why the first term of the series is considered apart.
$$\frac{d^\alpha}{dx^\alpha}\left(E_\alpha(x^\alpha)-1\right)=\frac{1}{\Gamma(-\alpha)}\sum_{k=1}^\infty \frac{1}{\Gamma(\alpha k+1)}\frac{d^\alpha }{dx^\alpha}(x^{\alpha k})$$ 
$\frac{d^\alpha }{dx^\alpha}(x^{\alpha k})=\frac{\Gamma(\alpha k+1)}{\Gamma\left(\alpha (k-1)+1\right)}x^{\alpha(k-1)}$
$$\frac{d^\alpha}{dx^\alpha}\left(E_\alpha(x^\alpha)-1\right)=\frac{1}{\Gamma(-\alpha)}\sum_{k=1}^\infty \frac{1}{\Gamma(\alpha k+1)}\frac{\Gamma(\alpha k+1)}{\Gamma\left(\alpha (k-1)+1\right)}x^{\alpha(k-1)}$$
$$\frac{d^\alpha}{dx^\alpha}\left(E_\alpha(x^\alpha)-1\right)=\frac{1}{\Gamma(-\alpha)}\sum_{k=1}^\infty \frac{x^{\alpha(k-1)}}{\Gamma\left(\alpha (k-1)+1\right)}=\frac{1}{\Gamma(-\alpha)}\sum_{h=0}^\infty \frac{x^{\alpha h}}{\Gamma\left(\alpha h+1\right)}$$
$$\frac{d^\alpha}{dx^\alpha}\left(E_\alpha(x^\alpha)-1\right)=E_\alpha(x^\alpha)$$
$$\frac{d^\alpha}{dx^\alpha}E_\alpha(x^\alpha)=E_\alpha(x^\alpha)+\frac{d^\alpha}{dx^\alpha}(1)$$
This is close to the expected equation 
$$\quad \frac{d^\alpha}{dx^\alpha}f(x)=f(x)\qquad \text{with} \quad f(x)=E_\alpha(x^\alpha)$$
But there is an extra term $\frac{d^\alpha}{dx^\alpha}(1)=\frac{x^{-\alpha}}{\Gamma(1-\alpha)}$
This is the difference compared to the case $\alpha=1$ of the exponential :
$$\frac{d^1}{dx^1}e^x=e^x+\frac{d^1}{dx^1}(1)=e^x$$
The first term in the series expansion of $e^x$ is constant$=1$. So its derivative is $0$, which is not the case for the fractional derivative of order different from $1$.
In fact, this difference comes from the definition of the lower bound $=0$ in the Riemann-Liouville operator for fractional differ-integration.
IN ADDITION :
In order to have a full solution, the Mittag-Leffler function has to be extended. Instead to limit the series to the terms with $k\geq 0$ consider all terms from $k=-\infty$ to $+\infty$.
$$f(x)=\sum_{k=-\infty}^\infty \frac{x^{\alpha k}}{\Gamma(\alpha k+1)}$$
The same calculus as above shows that $f(x)$ is a formal solution of the fractional differential equation 
$$\frac{d^\alpha}{dx^\alpha}f(x)=f(x)$$
Note :
Also, this is valid for the exponential function and $\alpha=1$ since $$\quad \frac{1}{k!}=\frac{1}{\Gamma(k+1)}=0 \quad\text{in}\quad k<0 \quad\to\quad e^x=\sum_{k=-\infty}^\infty \frac{x^k}{k!} $$ .
A: 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$.
A: 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 $k,n\in\mathbb{N}^*$
