Question: HI I am wondering if there is a unique matrix that maps $(x\_1,y\_1,z\_1)$ into $(x\_2,y\_2,z\_2)$. These two vectors have equal magnitude and are defined in orthogonal 3-D basis. If there is a unique solution how can I find it by considering rotation about all three orthogonal basis?

Answer: 
coffeemath explained why you cannot find an unique such matrix.

On another hand, since you are interested in rotations about the axes, there exists a pair of rotations $R\_1, R\_2$ about two of the axes of your choices, so that $v\_2=R\_2 R\_1 v\_1$. Moreover, if you fix the axes, and the order, I think there are exactly two such rotations.

To understand why, just think in spherical coordinates, where the angles are expressed with respect to the two axes you chose...

Then the first vector is $(R, \phi\_1, \theta\_1)$ and the second vector is 
$(R, \phi\_2, \theta\_2)$, so a rotation if angle $\phi\_1-\phi\_2$ and one of angle $\theta\_1-\theta\_2$ should do it.

The second possibility comes from the fact that you can also do a rotation of more than $180^\circ$ with respect to the second angle....
In the following, c stands for cosine and s for sine (of any angle $\theta$)
$$\pmatrix{c & s\\ -s &c} \pmatrix{x\_1 \\ y\_1}$$
So in particular, we have that $s^2 + c^2 = 1$.
The matrix multiply represents the rotation about the z-axis. The magnitude of the vector is preserved as such:
$$\left|\pmatrix{x\_1 \\ y\_1}\right|^2 = x\_1^2 + y\_1^2 \quad \text{this is the original magnitude}$$
$$\left|\pmatrix{cx\_1 +sy\_1\\ -sx\_1 + cy\_1}\right|^2 = (cx\_1 +sy\_1)^2 + (-sx\_1 + cy\_1)^2$$
$$=(c^2x\_1^2+s^2y\_1^2+2csx\_1y\_1) +(s^2x\_1^2+c^2y\_1^2-2csx\_1y\_1)$$
$$=c^2(x\_1^2+y\_1^2)+s^2(x\_1^2+y\_1^2)$$
$$=(c^2+s^2)(x\_1^2+y\_1^2)$$
$$=x\_1^2+y\_1^2 \quad \text{and the magnitude is unchanged}$$

You can apply this form of matrix a few times, starting with your first point in 3-D, and reach the desired value in one coordinate at a time. So for example, to rotate about the y axis:
$$\pmatrix{c & 0& s\\0&1&0\\-s&0&c}\pmatrix{x \\ y \\ z}$$
You would only need to do this twice, as the final value would by necessity match as desired, since the magnitudes are the same for the two points.

To find c and s, use $$c=\frac{x\_1}{\sqrt{x\_1^2+y\_1^2}} \quad\text{and}\quad s=\frac{y\_1}{\sqrt{x\_1^2+y\_1^2}}$$

HINT: any two values instead of $x\_1$ and $y\_1$ in that formula gives a valid c and s for a rotation matrix.