Is there a probability distribution with mean $1$ such that $f(x)=\frac{1}{x}f\left(\frac{1}{x}\right)$ Is there a probability distribution defined over $\mathbb{R}^{+}$ by the pdf $f$ such that, 
$$\forall x &gt; 0, f(x)=\frac{1}{x}f\left(\frac{1}{x}\right)$$
and
$$\int_0^{\infty} x~\mathrm{d}f = 1 $$

The first condition is easily satisfied by picking functions of the form

$$\left(\int_{-\infty}^{\infty} v(t)(x^t+x^{1-t})~\mathrm{d}t\right)^{-1}$$

However, in all cases, the first moment is always greater than one. Only by pushing the standard deviation to zero can it approach one. 

I realize the moment can be written as $$\int_0^1 \left(x+\frac{1}{x^2}\right)f(x)~\mathrm{d}x$$

and we have the constraint

$$\int_0^1 \left(1+\frac{1}{x}\right)f(x)~\mathrm{d}x=1$$

but I'm not sure if somehow that implies it has to be greater than one...
Since you already rewrote the moment and the density constraints, let's start from that.

We have $$1 = \int_0^1 \frac{x^3 + 1}{x^2}f(x)dx$$ and $$1 = \int_0^1 \frac{x^2 +x}{x^2}f(x)dx$$.

This gives $$0 = \int_0^1 (x^3 + 1 - x^2 - x) \frac{f(x)}{x^2}$$

But it is easy to see that $\forall x \in [0,1], x^3 + 1 - x^2 - x \geq 0$, hence we have a null integral with a non-negative integrand. This implies that $f(x) = 0$ almost everywhere, a contradiction.
The penultimate expression equals 1, which is correct. That implies that 

f(x) = (x+1/x^2)^-1, 

works.