A hyperbola passes through the foci of the ellipse $$\frac{x^2}{25} + \frac{y^2}{16} = 1$$ and its transverse and conjugate axes coincide with major and minor axes of the ellipse, respectively. If the product of their eccentricities is one, then the equation of the hyperbola is:

For the ellipse $$\frac{x^2}{25} + \frac{y^2}{16} = 1$$, we have $$a = 5$$, $$b = 4$$. The eccentricity is $$e_1 = \sqrt{1 - \frac{16}{25}} = \frac{3}{5}$$, and the foci are at $$(\pm 3, 0)$$.

Let the hyperbola be $$\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1$$ with eccentricity $$e_2$$. Since the product of eccentricities is 1, we have $$e_1 \cdot e_2 = 1$$, giving $$e_2 = \frac{5}{3}$$.

The hyperbola passes through the foci of the ellipse, so substituting $$(3, 0)$$: $$\frac{9}{A^2} = 1$$, giving $$A^2 = 9$$.

For the hyperbola, $$B^2 = A^2(e_2^2 - 1) = 9\left(\frac{25}{9} - 1\right) = 9 \cdot \frac{16}{9} = 16$$.

Therefore, the equation of the hyperbola is $$\frac{x^2}{9} - \frac{y^2}{16} = 1$$.