Assume I have a continuous random variable $X$ with pdf $f_X$. Then I have another random variable $Y = X/m$ for some $m \neq 0$. Consider $g:x \mapsto x/m$. Then $g^{-1}(x) = mx$. Hence $f_Y(y) = \left | \left ( g^{-1} \right)' (y) \right | f_X(g^{-1}(y)) = |m| f_X(my)$.
For $m < 0$, I have $$\int_\mathbb R f_Y (y) \, \mathrm{d}y = \int_\mathbb R |m| f_X(my) \, \mathrm{d}y = - \int_\mathbb R m f_X(my) \, \mathrm{d}y = - \int_\mathbb R f_X(x) \, \mathrm{d}x = -1.$$
Could you explain what's wrong with my computation?
This equality : $- \int_\mathbb R m f_X(my) \, \mathrm{d}y = - \int_\mathbb R f_X(x) $ is not true (obviously). When you do the change of variables $x = my$, the fact that $m < 0$ implies that the bounds of the integral should be reversed, which then implies multiplying the integral by $-1$.
If you want to do it rigorously just integrate over $[-t,t]$ :
$$ - \int_{-t}^t m f_X(my)dy = - \int_{-t}^t f_X(my)d(my) = - \int_{-t/m}^{t/m} f_X(x)dx $$
And finally, since $m<0$, the bounds need to be inverted and the integral has to be multiplied by $-1$, so :
$$ - \int_{-t}^t m f_X(my)dy = \int_{t/m}^{-t/m} f_X(x)dx $$
and then let $t \to \infty$ to get the correct result.