there is a lot of post about integral of the convolution of two fuction. Is it possible to demostrate that using variable changes as in the Fubini the following equivalence is valid:
\begin{align}\int_{-\infty}^{\infty} (h(x))(f*g)(x)dx&= \int_{-\infty}^{\infty} (f(x))(h*g)(x)dx \end{align}
Yes. I will leave out bounds of integration since we're working with the whole space anyway. Let $f,g,h : \mathbb{R}^n \to \mathbb{C}$ be positive or such that all integrals below converge absolutely. Then $$\begin{align*} \int f(x)(h*g)(x) dx &= \int f(x) \int h(y)g(x-y)\; dy dx \\ &= \iint f(x)g(x-y)h(y) \;dxdy \\ &= \int h(y) \int f(x)g(x-y)\; dx dy \\ &= \int h(y) \int f(u+y)g(u)\; du dy \\ &= \int h(y) (f*g)(-y) dy \end{align*}$$ where we used the change of variables $u = x-y$. Thus, unless you have some radial symmetry on either $h$ or $f*g$ (or, doing the argument the other direction, on $f$ or $h*g$), your desired identity does not necessarily hold.
Edit: You can of course also write out the other side directly: $$\begin{align*} \int h(y) (f*g)(y) dy &= \int h(y) \int f(x)g(y-x) dx dy \\ &= \iint f(x)g(y-x)h(y) dxdy \end{align*}$$ to notice that also radial symmetry on $g$ would suffice.