Two rotation cones have equal surface areas and equal volumes, with the radii of their base circles being $r_{1}$ and $r_{2},\left(\neq r_{1}\right)$. Express the common surface area, the common volume, and the radii of the spheres inscribed in each of the cones, respectively.
Let the heights of the cones be denoted by $m_{i}$, their slant heights by $a_{i}$ $(i=1,2)$, their common surface area by $F$, and their common volume by $V$. The known formulas for surface area and volume, along with the Pythagorean theorem, give us

$$
\begin{aligned}
F=\pi r_{i}\left(r_{i}+a_{i}\right), & (i=1,2) \\
V=\frac{\pi}{3} r_{i}^{2} m_{i}, & (i=1,2) \\
a_{i}^{2}=r_{i}^{2}+m_{i}^{2} . & (i=1,2)
\end{aligned}
$$

Since the left-hand sides of (1) and (2) are the same for $i=1$ and $i=2$, we can derive the system of equations for the unknowns $a_{1}, a_{2}$ using (3):

$$
\begin{aligned}
& r_{1}\left(r_{1}+a_{1}\right)=r_{2}\left(r_{2}+a_{2}\right) \\
& r_{1}^{4}\left(a_{1}^{2}-r_{1}^{2}\right)=r_{2}^{4}\left(a_{2}^{2}-r_{2}^{2}\right)
\end{aligned}
$$

Divide (5) by (4):

$$
r_{1}^{3}\left(a_{1}-r_{1}\right)=r_{2}^{3}\left(a_{2}-r_{2}\right)
$$

Then multiply (4) by $r_{1}^{2}$ and subtract (6) from the resulting equation:

$$
2 r_{1}^{4}=a_{2} r_{2}\left(r_{1}^{2}-r_{2}^{2}\right)+r_{2}^{2}\left(r_{1}^{2}+r_{2}^{2}\right)
$$

Rearranging, factoring, and dividing by $\left(r_{1}^{2}-r_{2}^{2}\right)$, we get

$$
a_{2} r_{2}=2 r_{1}^{2}+r_{2}^{2}
$$

Thus, from (1)-(3),

$$
\begin{gathered}
F=\pi\left(r_{2}^{2}+a_{2} r_{2}\right)=2 \pi\left(r_{1}^{2}+r_{2}^{2}\right) \\
V^{2}=\left(\frac{\pi}{3}\right)^{2} r_{2}^{2}\left(a_{2}^{2} r_{2}^{2}-r_{2}^{4}\right)=\left(\frac{\pi}{3}\right)^{2} r_{1}^{2} r_{2}^{2}\left(r_{1}^{2}+r_{2}^{2}\right) \\
V=\frac{2 \pi}{3} r_{1} r_{2} \sqrt{r_{1}^{2}+r_{2}^{2}}
\end{gathered}
$$

If, for example, we cut the first cone with a plane passing through its axis, the resulting isosceles triangle will have the circle inscribed in the sphere inscribed in the cone as its incircle, so the radius $\varrho_{1}$ of the sphere is equal to the ratio of the triangle's area to its semiperimeter:

$$
\varrho_{1}=\frac{r_{1} m_{1}}{r_{1}+a_{1}}=3 \frac{\frac{\pi}{3} r_{1}^{2} m_{1}}{\pi r_{1}\left(r_{1}+a_{1}\right)}=\frac{3 V}{F}=\frac{r_{1} r_{2}}{\sqrt{r_{1}^{2}+r_{2}^{2}}}
$$

From the derived $\varrho_{1}=\frac{3 V}{F}$, it is clear that the radius of the sphere inscribed in the second cone is the same.

Remarks. 1. In our solution, we divided by $r_{1}\left(r_{1}+a_{1}\right)$ and $\left(r_{1}^{2}-r_{2}^{2}\right)$: in our case, we know that both quantities are non-zero.

2. From the derived (7), the remaining dimensions of the cones can be easily calculated:

$$
a_{i}=2 \frac{r_{1}^{2}+r_{2}^{2}}{r_{i}}-r_{i}, \quad m_{i}=\frac{2 r_{1} r_{2} \sqrt{r_{1}^{2}+r_{2}^{2}}}{r_{i}^{2}} \quad(i=1,2)
$$

From these, it is clear that for any $r_{1} \neq r_{2}$ positive numbers, a pair of cones can be found that satisfy the conditions of the problem.