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: 
No.

For example, let $p\_1 = (x\_1,y\_1,z\_1)^T$, and $d\_1 = (x\_2,y\_2,z\_2)^T$. Choose $p\_2,p\_3 \in \mathbb{R}^3$ so that $p\_1, p\_2, p\_3$ are orthogonal. Let $d\_2, d\_3$ be arbitrary. Then define the matrix
$$A = \frac{1}{\|p\_1\|^2} d\_1 p\_1^T + d\_2 p\_2^T + d\_3 p\_3^T $$

It is easy to check that $A p\_1 = d\_1$. Since $d\_2, d\_3$ are arbitrary, it is clear that the transformation that maps $p\_1$ into $d\_1$ is not unique.

In fact, all transformations $A$ that satisfy $A p\_1 = d\_1$ can be expressed in this form with appropriate choice of $d\_2,d\_3$. If the $p\_k$ and $d\_k$ are chosen to be orthonormal, then the resulting $A$ will be a rotation (possibly improper).
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.