Five, try to find two different natural numbers, whose arithmetic mean $A$ and geometric mean $G$ are both two-digit numbers. Among $A, G$, one can be obtained by swapping the units and tens digits of the other.
Five, Solution Let these two natural numbers be $x_{1}, x_{2}$, then
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=2 A, \\
x_{1} x_{2}=G^{2} .
\end{array}\right.
$$

Thus, $x_{1}, x_{2}$ are the two roots of the equation $x^{2}-2 A x+G^{2}=0$. Therefore, $A \pm \sqrt{A^{2}-G^{2}}$ should be natural numbers, which means $\sqrt{A^{2}-G^{2}}$ must be a natural number.
Let $A=10 a+b \quad(1 \leqslant a, b \leqslant 9)$,

then $G=10 b+a$,
$$
A^{2}-G^{2}=9 \cdot 11 \cdot(a+b)(a-b) .
$$

For $A^{2}-G^{2}$ to be a perfect square, 11 must divide $a+b$ or $a-b$.
However, $1 \leqslant a-b \leqslant 8$,
so 11 cannot divide $a-b$.
Therefore, 11 must divide $a+b$.

Since $a+b \leqslant 5+8=17$, we know $a+b=11$, which requires $a-b$ to be a perfect square.

From $a-b=(a+b)-2 b=11-2 b$, we know $a-b$ must be a perfect square of an odd number.
But $a-b \leqslant 9-1=8<3^{2}$,
so $a-b=1^{2}=1$.
From $\left\{\begin{array}{l}a+b=11, \\ a-b=1\end{array}\right.$ we get $\left\{\begin{array}{l}a=6 \\ b=5 .\end{array}\right.$.

Thus, $A=65, G=56$.
$x_{1}, x_{2}$ are $65 \pm \sqrt{65^{2}-56^{2}}=65 \pm 33$.
Therefore, these two numbers are 98 and 32.