Let $f$ be a strictly positive function such that $\int\_{0}^{\infty}f(x)dx=\int\_{0}^{\infty}xf(x)=1$ (i.e., a probability density function with expectation one). Let also $g$ be a nonnegative nonconstant function which satisfies $\int\_{0}^{\infty}g(ax)f(x)dx=a$, $\forall a>0$. Does this imply that $g(x)=x$ a.e.?



Characterization of a particular integrable function
The answer is no. Let $h$ be a positive function on $R$ with the following properties:


a) Fourier transform has a pair of non-real zeros $\lambda, -\overline{\lambda}$,
symmetric with respect to the imaginary axis.


b) $\int\_{-\infty}^\infty h(t)dt=1,$


c) $\int\_{-\infty}^\infty e^t h(t)dt=1.$


Condition a) means 
$$\int\_{-\infty}^\infty e^{-i\lambda t}h(t)dt=0.$$
Multiplying this by $a^{-i\lambda}$, $a>0$ we obtain
$$\int\_{-\infty}^\infty e^{-i\lambda(t+\log a)}h(t)dt=0,$$
for every $a>0$. Making change of the variable $t=\log x$, we obtain
$$\int\_0^\infty e^{-i\lambda\log (ax)}h(\log x)\frac{dx}{x}=0.$$
Putting $f(x)=h(\log x)/x$, we obtain a function with the properties you stated;
they follow from b), c). Thus 
$$\int\_0^\infty (ax)^{-i\lambda} f(x)dx=0.$$
Now $g(x)=x+x^{-i\lambda}$ is a function $g$ which satisfies your identity.
If a positive function is required, we need $\lambda=\sigma-i$, and 
$$g(x)=x+k(x^{i\lambda}+x^{-i\overline{\lambda}})=x(1+k\cos(\sigma\log x)),$$
which is positive if $0<k<1$.


Existence of such $h$ is pretty evident. Take any probability density whose Fourier transform is analytic in some strip $|Im z|<c$ and has some zeros with negative imaginary part. Scaling will give you a zero whose imaginary part is $-1$. 
To achieve c), shift $h$,
replacing it with $h(t+c)$. This does not affect the zeros of Fourier transform.
Taking $h$ with Fourier transform having infinitely many zeros on a horizontal line, you obtain an
infinite dimensional space of $g$ with fixed $f$.


Edit. Sorry, I did not notice that you also need $g$ positive. I modified the example to achieve this additional property. 


