What function(s) can satisfy these conditions? For what function (or functions) is the following true:
1) $f(x)$ is positive for $x>0$
2) $\lim\limits_{x\to 0}{f(x)} = \infty$
3) $\lim\limits_{x\to\infty}f(x) = 0$
4) $\int_{0}^{\infty} {f(x)} dx = C$
5) $f(x)$ is symmetric over $y=x$
6) $f(x)$ isn't written in case structure
 A: A simple general recipe for these conditions is the implicit equation
$$g(x)g(y) = 1,$$
with $g$ being an increasing function such that $g(0) = 0$ and $g(x) \to \infty$ as $x \to \infty$. Then $f$ is given by
$$f(x) = g^{-1}\left(\frac 1 {g(x)}\right).$$
To get a finite integral for $\int _0^\infty f(x)$ under the given conditions, all you need is for $f(x)$ to decay faster than $1/x$ as $x \to \infty$. This is guaranteed as long as, informally, $g(x)$ grows "faster" towards $\infty$ as $x \to \infty$ than it decays to $0$ as $x \to 0$. (Formally, if $g(x)$ goes as $x^n$ as $x\to\infty$ and as $x^m$ as $x\to 0$, then you need $n > m$.)
You can recover Ilmari's example by taking
$$g(x) = \begin{cases}
x & \text{if }0 < x \le 1, \\
x^2 & \text{if }1 < x.
\end{cases}$$
For a smooth example, let $g(x) = e^x - 1$. Then you get
$$f(x) = -\ln\left(1 - e^{-x}\right),$$
for which WolframAlpha reports that $\int_0^\infty f(x) = \pi^2/6.$
A: As a particular example of a function satisfying Arturo Magidin's conditions,
$$f(x) = \begin{cases}
1/\sqrt x & \text{if }0 < x \le 1 \\
1/x^2 & \text{if }1 < x
\end{cases}$$
ought to work.  In particular, $\displaystyle \int_0^\infty f(x)\;dx = 3$ and $f(f(x)) = x$.

Addendum: If you don't want a piecewise defined function, you could write $f$ above as $f(x) = \exp\;g(\log x)$, where
$$g(u) = -\frac34 |u| - \frac54 u.$$
Of course, this is just a notational trick; $f$ still has a "kink" at $x=1$ due to the non-differentiability of $|u| = |\log x|$ there.  However, it's a trick that points in a useful direction: if you also want the function to be everywhere differentiable, you can replace $|u| = \sqrt{u^2}$ in the definition of $g$ above with the hyperbola $\sqrt{1+u^2}$ to get
$$\tilde g(u) = -\frac 34 \sqrt{1 + u^2} - \frac 54 u,$$
and thus
$$\tilde f(x) = \exp \left( -\frac 34 \sqrt{1 + (\log x)^2} - \frac 54 \log x \right).$$
This function $\tilde f$ still satisfies $\tilde f(\tilde f(x)) = x$, and since $\tilde f(x) < f(x)$ for all $x$, we know that its integral from 0 to infinity must be less than 3.  (Actually, it is about 2.19574343.) 
