A1, yes, the solution should be unique as long as the two sets of points are not collinear.
A2, your result matrix will be in the form:
[ A B C ]
[ D E F ]
C is your translation in x; F is your translation in y. Both are scale invariant.
If you take the left-most terms, these correspond to rotation and scale
[A B]
[D E]
in the following formula:
[I cos(t) -I sin(f)]
[J sin(t) J cos (F)]
where I is the scale in x and J is the scale in y.
To remove scaling, divide by A and B by sqrt(A*A+B*B) and D and E by sqrt(D*D+E*E)