Solution:
Let $A = |x^2 - 12x + 20|$.

The equation is:
$$\left(A^{\log x^2}\right)^{-1 + \log x} = A^{1 + \log(1/x)}$$

First, $A \geq 0$ and $x \neq 0$ (since $\log x^2$ is defined for $x \neq 0$).

Also, $\log x$ is defined for $x > 0$.

So, $x > 0$.

Now, $\log(1/x) = -\log x$, so $1 + \log(1/x) = 1 - \log x$.

The equation becomes:
$$\left(A^{\log x^2}\right)^{-1 + \log x} = A^{1 - \log x}$$

Recall that $A^{\log x^2} = A^{2 \log x} = (A^{\log x})^2$.

So,
$$\left((A^{\log x})^2\right)^{-1 + \log x} = A^{1 - \log x}$$
$$(A^{\log x})^{2(-1 + \log x)} = A^{1 - \log x}$$
$$A^{2(-1 + \log x) \log x} = A^{1 - \log x}$$

If $A = 0$, then $|x^2 - 12x + 20| = 0 \implies x^2 - 12x + 20 = 0 \implies x = 2$ or $x = 10$.

Check $x = 2$:
$A = 0$, $\log x = \log 2 > 0$, $\log x^2 = 2 \log 2 > 0$.

Left side: $0^{\text{something}}$ is $0$ if exponent $> 0$, $1$ if exponent $= 0$, undefined if exponent $< 0$.

But $A = 0$, so $0^{\text{something}}$ is $0$ if exponent $> 0$.

Right side: $0^{1 - \log 2}$. $1 - \log 2 < 1$, so exponent $< 1$.

But $0^a = 0$ for $a > 0$, undefined for $a \leq 0$.

So, $x = 2$ and $x = 10$ do not give valid solutions.

Now, for $A > 0$, $A \neq 0$:

Since $A > 0$, $A^k$ is defined for all real $k$.

So, equate exponents:

If $A \neq 1$, then $A^p = A^q \implies p = q$.

So,
$$2(-1 + \log x) \log x = 1 - \log x$$
Let $y = \log x$ (so $x > 0$):

$2(y - 1)y = 1 - y$

$2y^2 - 2y = 1 - y$

$2y^2 - 2y + y - 1 = 0$

$2y^2 - y - 1 = 0$

$2y^2 - y - 1 = 0$

Solve for $y$:

$y = \frac{1 \pm \sqrt{1 + 8}}{4} = \frac{1 \pm 3}{4}$

So $y = 1$ or $y = -\frac{1}{2}$.

So $\log x = 1 \implies x = 10$, $\log x = -\frac{1}{2} \implies x = 10^{-1/2} = 1/\sqrt{10}$.

Recall $A = |x^2 - 12x + 20|$.

If $A = 1$, then $A^p = A^q$ for any $p, q$.

So, $A = 1 \implies |x^2 - 12x + 20| = 1$.

$x^2 - 12x + 20 = 1$ or $x^2 - 12x + 20 = -1$

$x^2 - 12x + 19 = 0$

$x = 6 \pm \sqrt{36 - 19} = 6 \pm \sqrt{17}$

$x^2 - 12x + 21 = 0$

$x = 6 \pm \sqrt{36 - 21} = 6 \pm \sqrt{15}$

But $x > 0$.

So, possible $x$ values:
$6 + \sqrt{17}$, $6 - \sqrt{17}$, $6 + \sqrt{15}$, $6 - \sqrt{15}$.

Check if these are positive:
$\sqrt{17} \approx 4.12$, $6 - 4.12 = 1.88 > 0$
$\sqrt{15} \approx 3.87$, $6 - 3.87 = 2.13 > 0$

So all four are positive.

Now, check if $x = 10$ or $x = 1/\sqrt{10}$ gives $A = 1$.

For $x = 10$:
$A = |100 - 120 + 20| = |0| = 0 \neq 1$

For $x = 1/\sqrt{10}$:
$x^2 = 1/10$, $x^2 - 12x + 20 = 1/10 - 12/\sqrt{10} + 20$

This is not $\pm 1$.

So, $x = 10$ and $x = 1/\sqrt{10}$ are not among the $A = 1$ solutions.

Now, check if $A > 0$ for $x = 1/\sqrt{10}$:
$x = 1/\sqrt{10}$, $x^2 = 1/10$, $x^2 - 12x + 20 = 1/10 - 12/\sqrt{10} + 20$

$12/\sqrt{10} \approx 12/3.16 \approx 3.8$

So $1/10 - 3.8 + 20 \approx 0.1 - 3.8 + 20 = 16.3$

So $A > 0$.

So, $x = 1/\sqrt{10}$ is valid.

So, the solutions are:

- $x = 1/\sqrt{10}$
- $x = 6 + \sqrt{17}$
- $x = 6 - \sqrt{17}$
- $x = 6 + \sqrt{15}$
- $x = 6 - \sqrt{15}$

Total: $5$ real numbers $x$.

Answer: $\boxed{5}$