geometric hash code or is there a unique affine transformation mapping two 2D points onto (0,0) and (1,1)? How can I compute it? I have a 2D transformation $T$ composed by a scale $\lambda$, a rotation by angle $\theta$ and a translation vector $\begin{bmatrix}t_x\\t_y\end{bmatrix}$:
$$
    T=\begin{bmatrix}
    \lambda\cos(\theta) &     -\lambda\sin(\theta) & t_x \\
    \lambda\sin(\theta) & \lambda\cos(\theta) & t_y \\
    0 & 0 & 1 \\
    \end{bmatrix}
$$
$T$ operates on 2D point $P$ expressed with homogeneous coordinates:
$$
P=\begin{bmatrix}
x\\
y\\
1
\end{bmatrix}
$$
$T$ maps 2D $(x,y)$ point to 2D $(u,v)$ point according to:
$$
\begin{bmatrix}u\\v\\1\end{bmatrix}=TP
$$
Now I have 
$$
A=\begin{bmatrix}x_A\\y_A\\1\end{bmatrix},B=\begin{bmatrix}x_B\\y_B\\1\end{bmatrix}
$$
I would like to map $A$ to $(0,0)$ and $B$ to $(1,1)$:
$$\begin{bmatrix}0\\0\\1\end{bmatrix}=TA$$
$$\begin{bmatrix}1\\1\\1\end{bmatrix}=TB$$
Is $T$ unique?
How can I compute $T$?

Background information:

I am trying to compute an hash code as described in 
LANG, Dustin, et al. Astrometry.net: Blind astrometric calibration of arbitrary astronomical images. The astronomical journal, 2010, 139.5: 1782.
See here Figure 1 from the above paper:


Partial solution
$ \begin{cases} 
\begin{align*}
0 &= x_{A}\lambda \cos(\theta) - y_{A}\lambda \sin(\theta) + t_{x} \\
0 &= x_{A}\lambda \sin(\theta) + y_{A}\lambda \cos(\theta) + t_{y} \\
0 &= x_{B}\lambda \cos(\theta) - y_{B}\lambda \sin(\theta) + t_{x} - 1\\
0 &= x_{B}\lambda \sin(\theta) + y_{B}\lambda \cos(\theta) + t_{y} - 1\\
\end{align*} 
\end{cases} $
Jacobian:
$J=\begin{bmatrix}
x_{A}\cos(\theta) - y_{A} \sin(\theta) & 
-x_{A}\lambda \sin(\theta) - y_{A}\lambda \cos(\theta) &
1 &
0 \\
x_{A}\sin(\theta) + y_{A}\cos(\theta) &
x_{A}\lambda \cos(\theta) - y_{A}\lambda \sin(\theta) &
0 &
1 \\
x_{B}\cos(\theta) - y_{B} \sin(\theta) & 
-x_{B}\lambda \sin(\theta) - y_{B}\lambda \cos(\theta) &
1 &
0 \\
x_{B}\sin(\theta) + y_{B}\cos(\theta) &
x_{B}\lambda \cos(\theta) - y_{B}\lambda \sin(\theta) &
0 &
1 \\
\end{bmatrix}$
The rows of J are independent unless:


*

*$\lambda = 0$ or 

*$x_{A} = x_{B}$ and $y_{A} = y_{B}$.


In that case, $J$ is invertible. 
However, that doesn't guarantee that there is only one solution:
Unique solution to system of nonlinear equations (non-singular Jacobian)
On the other hand, if $\lambda = 0$ is a solution, then the solution is not unique. I can linearize the system around the solution. Since the null space of the Jacobian is non-trivial, the solution is not unique.
That means if a translation can translate points A and B to $(0, 0)$ and $(1, 1)$, then the solution (in terms of $\lambda, \theta, t_{x}, t_{y}$)  is not unique. But in that case, $T$ is still unique.
