Why does an affine transformation $A$ when constrained by $A^TA=\lambda^2I$ result in a similarity transformation? Why does an affine transformation $A$ when constrained by $A^TA=\lambda^2I$ result in a similarity transformation?
I came across this when studying linear transformations in these notes which says:

The similarity group is obtained from the affine group by requiring that $A$ be orthogonal: $A^TA=\lambda^2I$

Can't seem to wrap my head around this one.
 A: Let $\mathbf {x, y} \in \mathbb{R}^n$ be unit vectors.
Then, the angle $\theta_1$ between them is given by $ \cos \theta_1 = \mathbf{x^\top y}$.
Also, angle between $\mathbf{Ax}$ and $\mathbf{Ay}$ is given by :
$\cos \theta_2 = \dfrac{(\mathbf{Ax})^\top (\mathbf{Ay})}{\|\mathbf{Ax}\|\|\mathbf{Ay}\|}$.
Now, $\|\mathbf{Ax}\|_2^2 = (\mathbf{Ax})^\top(\mathbf{Ax}) = \mathbf{x^\top A^\top Ax} = \mathbf{\lambda ^2 x^\top x}$.
Thus, $\mathbf{\|Ax\|} = \mathbf{|\lambda|}$.
$\implies \cos \theta_2 = \mathbf{\dfrac{\lambda^2 x^ \top y}{\lambda^2}} = \mathbf{x^ \top y} \implies \theta _1 = \theta _2$.
Thus, $\mathbf{A}$ preserves angles and is a similarity transformation.
A: Orthogonality is forced by requiring a scalar multiple of the identity; if any two columns $a_i, a_j$ were not orthogonal, there would be a nonzero entry in the corresponding entries $b_{ij}, b_{ji}$ of the product.
Orthogonality results in the absence of any shear/twist in the transformation, restricting it to only reflection, rotation and translation.
The eigenvalue squares as scalars allow for non-normality, as it is possible to scale the original space without affecting the angles between lines in it, provided all axes of the space are scaled by the same amount. If $A$ was instead orthonormal, you force a lack of scaling as well.
See also this question, for which answers explain a converse point, why similarity transformation is a subtype of affine transformation.
A: Let $A=RU$ be the polar decomposition of $A$, where $R$ is positive semi-definite and $U$ is unitary.
Then $R^2=A^TA=\lambda^2I$; being diagonalizable  this forces $R=|\lambda|I$ and thus $A=|\lambda|U$, a similarity transformation.
