Let $10a + b$ be the arithmetic mean of the given numbers, where $a$ and $b$ are decimal digits. Let $10a + b + x$ and $10a + b - x$ be the numbers we are searching for. Then, by the premises
$$\sqrt{(10a + b + x)(10a + b - x)} = 10b + a,$$
which after squaring and simplifying gives $x^2 = 99(a^2 - b^2)$. So, $x^2$ is divisible by $99$, implying $x^2$ is divisible by $3$ and $11$. Since $3$ and $11$ are primes, $x$ itself is divisible by $3$ and $11$, and therefore by $33$. Denoting $x = 33z$, we get
$$11z^2 = \frac{99 \cdot 11z^2}{99} = \frac{x^2}{99} = a^2 - b^2 = (a+b)(a-b),$$
from which we see, that the product $(a+b)(a-b)$ is divisible by $11$. Since $11$ is a prime, either $a+b$ or $a-b$ is divisible by $11$. Since $a-b \neq 0$ and $a$ and $b$ are single-digit numbers, we must have $a+b = 11$. Therefore $a-b = z^2$. Since $a-b$ and $a+b$ are either both odd or both even, $z$ must be odd. So, $z=1$, since $z \ge 3$ implies $x \ge 99$, but $10a+b-x$ must be positive. So, $a=6$, $b=5$, $x=33$, and the corresponding pair is $(98, 32)$.