Sorry to give yet another answer that does not address the issue of fractional $n$ [it seems that fractional derivatives are not such a familiar topic to many research mathematicians; certainly they're not to me], but:
There is a little issue here which has not been addressed. By the context of the OP's question, I gather s/he is looking for real-valued functions which are equal to their $n$th derivative (and not their $k$th derivative for $k < n$). Several answerers have mentioned that the set of solutions to $f^{n} = 0$ forms an $n$-dimensional vector space. But over what field? It is easier to identify the space of such complex-valued functions, i.e., $f: \mathbb{R} \rightarrow \mathbb{C}$: namely, a $\mathbb{C}$-basis is given by $f(x) = e^{2 \pi i k/n}$ for $0 \leq k < n$. But what does this tell us about the $\mathbb{R}$-vector space of real-valued solutions to this differential equation?
The answer is that it is $n$-dimensional as a $\mathbb{R}$-vector space, though it does not have such an immediately obvious and nice basis.
Let $W$ be the $\mathbb{R}$-vector space of real-valued functions $f$ with $f^{(n)} = 0$ and $V$ the $\mathbb{C}$-vector space of $\mathbb{C}$-valued functions $f$ with $f^{(n)} = 0$.
There is a natural inclusion map $W \mapsto V$. Phrased algebraically, the question is whether the induced map $L: W \otimes_{\mathbb{R}} \mathbb{C} \rightarrow V$ is an isomorphism of $\mathbb{C}$-vector spaces. In other words, this means that any given $\mathbb{R}$-basis of $W$ is also a $\mathbb{C}$-basis of $V$. This is certainly not automatic. For instance, viewing the Euclidean plane as first $\mathbb{R}^2$ and second as $\mathbb{C}$ gives a map
$\mathbb{R}^2 \rightarrow \mathbb{C}$ which certainly does not induce an isomorphism upon tensoring with $\mathbb{C}$, since the first space has (real) dimension $2$ but the second space has (complex) dimension $1$.
For more on this, see Theorem 1.6 of
http://www.math.uconn.edu/~kconrad/blurbs/galoistheory/galoisdescent.pdf
It turns out that this is actually a problem in Galois descent: according to Theorem 2.14 of the notes of Keith Conrad already cited above, the map $L$ is an isomorphism iff there exists a conjugate-linear involution $r: V \rightarrow V$, i.e., i.e., a map which is self-inverse and satisfies, for any $z \in \mathbb{C}$ and $v \in V$, $r(zc) = \overline{z} r(c)$.
But indeed we have such a thing: an element of $V$ is just a complex-valued function $f$,
so we put $r(f) = \overline{f}$. Note that this stabilizes $V$ since the differential equation $f^{(n)}) = 0$ "is defined over $\mathbb{R}$": or more simply, the complex conjugate of the $n$th derivative is the $n$th derivative of the complex conjugate. Thus we have "descent data" (or, in Keith Conrad's terminology, a G-structure) and the real solution space has the same dimension as the complex solution space.
It is a nice exercise to use these ideas to construct an explicit real basis of $W$.