As I know, a rigid body is a set of $N$ particles in three-dimensional space subject to the following constraint: if $b_1,\dots,b_N\in \mathbb{R}^3$ are the initial positions of the particles and if $c_1,\dots,c_N: \mathbb{R}\to \mathbb{R}^3$ are their time evolutions, then
$$|c_{i}(t)-c_{j}(t)|=|b_i-b_j| , \quad \forall t \in \mathbb{R}.$$
In other words, the distance between each pair of points is maintained constant throughout the time evolution. With this in mind, it is usual to consider that the configuration manifold of a rigid body is in truth
$$Q = \mathbb{R}^3\times SO(3).$$
In Spivak's Mechanics book it is said:
This all can be expressed in a more familiar, geometric, way by considering the "configuration space" of $\mathbf{b}$, which is the subset $\mathscr{M}\subset (\mathbb{R}^3)^N$ of all points that can be reached from $\mathbf{b}$ at the end of a rigid motion. In other words,
$$\mathscr{M}=\{(A(b_1),\dots,A(b_N)) : A \ \text{an orientation preserving isometry of } \ \mathbb{R}^3\}.$$
When $\mathbf{b}$ is non-planar, $\mathscr{M}$ is a $6$-dimensional manifold diffeomorphic to the set of all orientation preserving isometries $A$ of $\mathbb{R}^3$, and thus to $\mathbb{R}^3\times SO(3)$. With this picture, a rigid motion of $\mathbf{b}$is simply a curve in $\mathscr{M}$, so a virtual infinitesimal displacement $\mathbf{v}$ of $\mathbf{b}$ is simply a tangent vector to $\mathscr{M}$ at $\mathbf{b}$.
Although this seems really simple, I'm having a hard time to understand that the configuration manifold of a rigid body is $\mathbb{R}^3\times SO(3)$. I see that in truth if $A$ is a curve of isometries, then really $t\mapsto (A(t)(b_1),\dots,A(t)(b_N))$ is a curve that satisfies the constraints.
But why this says the configuration manifold is the set of all orientation preserving isometries? Also, where the orientation preserving came from? In short, how to see that the configuration manifold of a rigid body is $\mathbb{R}^3\times SO(3)$?