Is there a vector field that is equal to its own curl? I was wondering if there is a vector field that satisfies the following condition: 
$$\vec F=\nabla \times \vec F$$
 A: Jesse Madnick's answer notwithstanding one might be interested in nontrivial explicit solutions from which more complicated solutions can be generated by transportation and superposition.
I'm looking for solutions that do not depend on the third variable – "plane waves", so to speak. When $F=(F_1,F_2,F_3)$ is such a solution then
$${\rm curl}\,F=(F_{3.2},-F_{3.1}, F_{2.1}-F_{1.2})\ .$$
We write $F_3=:u$. The condition ${\rm curl}\,F=F$ then leads to
$F_1=u_y$, $F_2=-u_x$, and above all to 
$$u+\Delta u=0\ .\tag{1}$$
The well known equation $(1)$ can be treated using separation of variables. From a geometrical standpoint it would be natural to try $u=R(r)\>\Phi(\phi)$, but this leads to Bessel's differential equation $r^2 R''+r R'+(r^2-k^2)R=0$, $\>k\in{\mathbb N}$, whose solutions are non elementary.
When we write $u=X(x)\>Y(y)$ instead we are lead to
$$X''+\lambda X=0,\qquad Y''+(1-\lambda) Y=0\ ;$$
here $\lambda\in{\mathbb R}$ is arbitrary. When $0<\lambda<1$ we obtain solutions $u$ which are periodic in $x$ and in $y$. For other values of $\lambda$ the resulting $u$ is periodic in one variable, and a hyperbolic function in the other. The values $\lambda=0$ and $\lambda=1$ are special; here one of $X(\cdot)$ and $Y(\cdot)$ is linear, the other periodic with period $2\pi$.
A: I'd like to offer a high-brow answer to this question.  While I realize that this solution is likely to be above the OP's level, I'm hoping it'll be of some interest to others.

Claim: Let $u(x,y)$, $v(x,y)$, $w(x)$ be any real-analytic functions (defined on a neighborhood of $0$).  Then there exists a unique real-analytic vector field $F = (F^1, F^2, F^3)$ that satisfies $\text{curl}\,F = F$ and
  $$\begin{align}
F^1(x,0,0) & = w(x) \\
F^2(x,y,0) & = u(x,y) \\
F^3(x,y,0) & = v(x,y).
\end{align}$$

We'll show this by applying the Cauchy-Kovalevskaya Theorem twice.  Here I should acknowledge that I first saw this techinque applied to $\text{curl}\,F = F$ in an exercise in Robert Bryant's "Nine Lectures on Exterior Differential Systems."
Cauchy-Kovalevskaya Theorem: If $\mathbf{H}$ and $\phi$ are real-analytic functions near the origin, then there is a neighborhood of the origin on which there exists a unique real-analytic solution $\mathbf{g}(\mathbf{x}, t)$ to
$$\begin{align}
\frac{\partial \mathbf{g}}{\partial t}(\mathbf{x},t) & = \mathbf{H}(\mathbf{x},t, \mathbf{g}(\mathbf{x},t), \frac{\partial \mathbf{g}}{\partial x^i}) \\
\mathbf{g}(\mathbf{x}, 0) & = \phi(\mathbf{x})
\end{align}$$
A PDE system of this form will be called a "Cauchy problem."

Setup
The equation $\text{curl}\,F = F$ can be viewed as an overdetermined system of PDEs -- in fact, a system of 4 first-order quasilinear equations for 3 unknown functions.  That is, if we write $F = (F^1, F^2, F^3)$, then the condition $\text{curl}\,F = F$ becomes:
$$\begin{align}
\frac{\partial F^2}{\partial z} - \frac{\partial F^3}{\partial y} & = F^1 \tag{1} \\
\frac{\partial F^3}{\partial x} - \frac{\partial F^1}{\partial z} & = F^2 \tag{2} \\
\frac{\partial F^1}{\partial y} - \frac{\partial F^2}{\partial x} & = F^3. \tag{3}
\end{align}$$
The fourth equation is a hidden "integrability condition" of sorts: namely, any solution $F$ to $\text{curl}\,F = F$ must satisfy $\text{div}\,F = 0$.  This gives us a fourth (first-order quasilinear) equation
$$\frac{\partial F^1}{\partial x} + \frac{\partial F^2}{\partial y} + \frac{\partial F^3}{\partial z} = 0. \tag{4}$$

The Cauchy Problems
We can write this system as a sequence of two Cauchy problems as follows.  Let $u(x,y)$, $v(x,y)$, and $w(x)$ be arbitrary real-analytic functions.
We first consider the problem of finding $g(x,y)$ such that
$$\begin{align}
\frac{\partial g}{\partial y} & = \frac{\partial u}{\partial x} + v(x,y) \\
g(x,0) & = w(x).
\end{align}$$
By Cauchy-Kovalevskaya$^\dagger$, there exists a unique real-analytic solution $g(x,y)$.
Second, letting $g(x,y)$ be a solution as above, we consider the problem of finding $F^1, F^2, F^3$ such that
$$\begin{align}
\frac{\partial F^1}{\partial z} & = -F^2 + \frac{\partial F^3}{\partial x} \\
\frac{\partial F^2}{\partial z} & = F^1 + \frac{\partial F^3}{\partial y} \\
\frac{\partial F^3}{\partial z} & = -\frac{\partial F^1}{\partial x} - \frac{\partial F^2}{\partial y} \\
F^1(x,y,0) & = g(x,y) \\
F^2(x,y,0) & = u(x,y) \\
F^3(x,y,0) & = v(x,y)
\end{align}$$
By Cauchy-Kovalevskaya again, there exists a unique real-analytic solution $F = (F^1, F^2, F^3)$.
By construction, this solution $(F^1, F^2, F^3)$ satisfies equations (1), (2) and (4).  One can check (exercise!$^*$) that this $F$ necessarily satisfies equation (3) as well.  This completes the proof. $\lozenge$

End notes
$^\dagger$ Admittedly overkill here
$^*$ Hint for Exercise: Consider $E(x,y,z) := \frac{\partial F^1}{\partial y} - \frac{\partial F^2}{\partial x} - F^3$.  Check that $E(x,y,0) = 0$ and $\frac{\partial E}{\partial z} = 0$, and then apply uniqueness in Cauchy-Kovalevskaya.
Note that this exercise shows why we needed the first Cauchy problem at all: that is, we couldn't just choose $F^1(x,y,0) = g(x,y)$ completely arbitrarily!
A: To give a final answer, I've found that, in the standard unit coordinates, the field $$F(x,y,z) = (-\cos z, \sin z, 0)$$
is equal to its curl. Again, brought to you by this paper.
P.S. My first attempt revealed that Wolfram Alpha is a wise-guy.
A: if $f =(\sin z, \cos z, 0)\ $  then $\nabla \times f = f$
