1. Given condition

We have $a^2+b^2+c^2=1$ and $a\\cos\theta=b\\cos\left(\theta+\frac{2\pi}{3}\right)=c\\cos\left(\theta+\frac{4\pi}{3}\right),$ with $\theta=\frac{\pi}{9}.$

Let the common value be $k$. Then $a\cos\theta=k,\qquad b\cos\left(\theta+\frac{2\pi}{3}\right)=k,\qquad c\cos\left(\theta+\frac{4\pi}{3}\right)=k.$ So, $a=\frac{k}{\cos\theta},\qquad b=\frac{k}{\cos\left(\theta+\frac{2\pi}{3}\right)},\qquad c=\frac{k}{\cos\left(\theta+\frac{4\pi}{3}\right)}.$

2. Write the two vectors

Let $\vec u=(a,b,c),\qquad \vec v=(b,c,a).$ We need the angle $\phi$ between them.

Using dot product, $\cos\phi=\frac{\vec u\cdot \vec v}{|\vec u||\vec v|}.$

Now, $\vec u\cdot \vec v=ab+bc+ca.$ Also,

and similarly $|\vec v|=\sqrt{b^2+c^2+a^2}=1.$ Hence, $\cos\phi=ab+bc+ca.$ So it is enough to compute $ab+bc+ca$.

3. Express in terms of secants

Let $\alpha=\theta=\frac{\pi}{9},\qquad \beta=\theta+\frac{2\pi}{3},\qquad \gamma=\theta+\frac{4\pi}{3}.$ Then $a=k\sec\alpha,\quad b=k\sec\beta,\quad c=k\sec\gamma.$ Therefore, $ab+bc+ca=k^2\big(\sec\alpha\sec\beta+\sec\beta\sec\gamma+\sec\gamma\sec\alpha\big).$

Also from $a^2+b^2+c^2=1$, $k^2\left(\sec^2\alpha+\sec^2\beta+\sec^2\gamma\right)=1.$

So we need relations among these trig terms.

4. Use the identity for angles differing by $\frac{2\pi}{3}$

Let $x=\cos\alpha,\quad y=\cos\beta,\quad z=\cos\gamma.$ Since $\alpha+\beta+\gamma=3\theta+2\pi,$ and here $3\theta=\frac{\pi}{3}$, these are the three cosine values at angles separated by $\frac{2\pi}{3}$.

A standard identity is $\cos\alpha+\cos\left(\alpha+\frac{2\pi}{3}\right)+\cos\left(\alpha+\frac{4\pi}{3}\right)=0,$ so $x+y+z=0.$

Now divide by $xyz$: $\frac{x+y+z}{xyz}=0$ which gives $\frac{1}{yz}+\frac{1}{zx}+\frac{1}{xy}=0.$ That is, $\sec\beta\sec\gamma+\sec\gamma\sec\alpha+\sec\alpha\sec\beta=0.$

Hence, $ab+bc+ca=k^2\cdot 0=0.$

Therefore, $\cos\phi=0\implies \phi=\frac{\pi}{2}.$

5. Conclusion

The angle between the vectors is $\boxed{\frac{\pi}{2}}.$

So the correct option is D.