Let |\vec a|=a, \qquad |\vec b|=b and the angle between them be $\theta$, where $\cos\theta=\frac{5}{9}.$

Use the magnitude formulas: |\vec a+\vec b|^2=a^2+b^2+2ab\cos\theta, |\vec a-\vec b|^2=a^2+b^2-2ab\cos\theta.

Given |\vec a+\vec b|=\sqrt{2}\vec a-\vec b, so on squaring both sides, |\vec a+\vec b|^2=2|\vec a-\vec b|^2.

Substitute the formulas: $a^2+b^2+2ab\cos\theta=2(a^2+b^2-2ab\cos\theta).$

Now put $\cos\theta=\frac{5}{9}$: $a^2+b^2+\frac{10}{9}ab=2a^2+2b^2-\frac{20}{9}ab.$

Bring all terms to one side: $0=a^2+b^2-\frac{30}{9}ab=a^2+b^2-\frac{10}{3}ab.$

So, $a^2+b^2=\frac{10}{3}ab.$

Given $a=nb$, substitute: $(nb)^2+b^2=\frac{10}{3}(nb)(b).$

$n^2b^2+b^2=\frac{10}{3}nb^2.$

Divide by $b^2$: $n^2+1=\frac{10}{3}n.$

Solve the quadratic: $3n^2-10n+3=0.$

Factorize: $3n^2-9n-n+3=0$ $3n(n-3)-1(n-3)=0$ $(3n-1)(n-3)=0.$

Hence, $n=\frac13 \quad \text{or} \quad n=3.$

Since the question asks for the integer value of $n$, the only integer solution is $\boxed{3}.$