To solve this problem, we will use the concepts of vector addition and magnitudes involving the dot product. Given the angle between $$\vec{a}$$ and $$\vec{b}$$ is $$\cos^{-1}\left(\frac{5}{9}\right)$$, we can use the properties of dot products and magnitudes to find the required integer value of $$n$$.

First, let's use the condition $$|\vec{a} + \vec{b}| = \sqrt{2} |\vec{a} - \vec{b}|$$. Square both sides to remove the square roots:

$$ |\vec{a} + \vec{b}|^2 = 2 |\vec{a} - \vec{b}|^2 $$

Now we will expand both sides using the formula for the magnitude of the sum and difference of vectors:

$$ |\vec{a} + \vec{b}|^2 = (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = \vec{a} \cdot \vec{a} + 2 \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b} $$

And

$$ |\vec{a} - \vec{b}|^2 = (\vec{a} - \vec{b}) \cdot (\vec{a} - \vec{b}) = \vec{a} \cdot \vec{a} - 2 \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b} $$

Substitute these back into the original equation:

$$ \vec{a} \cdot \vec{a} + 2 \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b} = 2 (\vec{a} \cdot \vec{a} - 2 \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b}) $$

Expand the right-hand side:

$$ \vec{a} \cdot \vec{a} + 2 \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b} = 2 \vec{a} \cdot \vec{a} - 4 \vec{a} \cdot \vec{b} + 2 \vec{b} \cdot \vec{b} $$

Rearrange all terms to one side to combine like terms:

$$ \vec{a} \cdot \vec{a} + 2 \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b} - 2 \vec{a} \cdot \vec{a} + 4 \vec{a} \cdot \vec{b} - 2 \vec{b} \cdot \vec{b} = 0 $$

Combine like terms:

$$ - \vec{a} \cdot \vec{a} + 6 \vec{a} \cdot \vec{b} - \vec{b} \cdot \vec{b} = 0 $$

We know that:

$$ \vec{a} \cdot \vec{a} = |\vec{a}|^2 \quad \text{and} \quad \vec{b} \cdot \vec{b} = |\vec{b}|^2 \quad \text{and} \quad \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos(\theta) $$

Since the angle between $$\vec{a}$$ and $$\vec{b}$$ is $$\cos^{-1}\left(\frac{5}{9}\right)$$, we have:

$$ \cos(\theta) = \frac{5}{9} $$

Substitute these back into the equation:

$$ -|\vec{a}|^2 + 6 |\vec{a}| |\vec{b}| \left(\frac{5}{9}\right) - |\vec{b}|^2 = 0 $$

Simplify it further:

$$ -|\vec{a}|^2 + \frac{30}{9} |\vec{a}| |\vec{b}| - |\vec{b}|^2 = 0 $$

$$ -|\vec{a}|^2 + \frac{10}{3} |\vec{a}| |\vec{b}| - |\vec{b}|^2 = 0 $$

We know that $$|\vec{a}| = n |\vec{b}|$$. Substitute this into the equation:

$$ -(n |\vec{b}|)^2 + \frac{10}{3} (n |\vec{b}|) |\vec{b}| - |\vec{b}|^2 = 0 $$

Simplify it:

$$ -n^2 |\vec{b}|^2 + \frac{10}{3} n |\vec{b}|^2 - |\vec{b}|^2 = 0 $$

Factor out $$|\vec{b}|^2$$:

$$ |\vec{b}|^2 \left(-n^2 + \frac{10}{3} n - 1\right) = 0 $$

Since $$|\vec{b}|^2 \neq 0$$, we can solve:

$$ -n^2 + \frac{10}{3} n - 1 = 0 $$

Multiply through by 3 to clear the fraction:

$$ -3n^2 + 10n - 3 = 0 $$

Rearrange it to match standard quadratic form:

$$ 3n^2 - 10n + 3 = 0 $$

Solve this quadratic equation using the quadratic formula:

$$ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$

Here, $$a = 3$$, $$b = -10$$, and $$c = 3$$:

$$ n = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 3 \cdot 3}}{2 \cdot 3} = \frac{10 \pm \sqrt{100 - 36}}{6} = \frac{10 \pm \sqrt{64}}{6} = \frac{10 \pm 8}{6} $$

This results in two possible solutions for $$n$$:

$$ n = \frac{18}{6} = 3 \quad \text{and} \quad n = \frac{2}{6} = \frac{1}{3} $$

However, since $$n$$ is given to be an integer, we take:

$$ \boxed{3} $$