Geometric interpretation of the elements of $SO(3)$ It is well know that $SO(3)$ is defined as:
$$O(3)=\big\{\psi\in GL(3,\mathbb R):\psi^t\circ\psi=id\big\}$$
What is the geometric idea behind $\psi\circ\psi^t=id$?
Maybe, if we impose that a linear map $\psi:\mathbb R^3\to \mathbb R^3$ is bijective and has a one fixed point, could we deduce that $\psi^t\circ\psi=id$?
 A: The idea is that the orthogonal matrices preserve distance and angle. The dot product (scalar product) give us distances and angles. 
For example: $\langle \bf v, \bf v \rangle = \| \bf v \|^2$ where $\langle \bf v, \bf v\rangle$ is the dot product (scalar product) of $\bf v$ with $\bf v$, and $\| \bf v \|$ is the length of $\bf v$. While $\langle  \bf v, \bf w \rangle = \|\bf v\|\|\bf w\|\cos\theta$, where $\theta$ is the included angle between $\bf v$ and $\bf w$.
Given a column vectors
$${\bf v} = \left(\begin{array}{c} 1 \\ 2 \\ 3\end{array}\right)
\ \ \ \ \ {\bf w} = \left(\begin{array}{c} 2 \\ 5 \\ 9\end{array}\right)$$
We can find the dot product (scalar product) by matrix multiplication: $\langle \bf v, \bf w \rangle = {\bf v}^{\top}{\bf w}$
$$\left(\begin{array}{ccc} 1 & 2 & 3 \end{array}\right)\left(\begin{array}{c} 2 \\ 5 \\ 9\end{array}\right) = 1\times 2+2\times 5+3\times 9 = 39$$
The dot product (scalar product) can be given by $\langle \bf v, \bf w \rangle = \bf v^{\top}\bf w$.
Now let's say we perform a linear transformation. Let $M$ be the matrix transformation. We have $\bf v$ sent to $M\bf v$ and $\bf w$ set to $M\bf w$. 
What happens to the angles and the distances? What happens to $\langle \bf v, \bf w \rangle$? Well
$$\langle {\bf v}, {\bf w} \rangle \longmapsto \langle M{\bf v}, M\bf w \rangle$$
Using matrix multiplication we have
\begin{eqnarray*}
\langle M{\bf v}, M\bf w \rangle &=& (M{\bf v})^{\top}(M{\bf w}) \\ 
&=& {\bf v}^{\top}M^{\top}M{\bf w}
\end{eqnarray*}
The dot product (scalar product), which measures angles and distances, is preserved if, and only if,
\begin{eqnarray*}
\langle {\bf v}, {\bf w}\rangle &=& \langle M{\bf v}, M{\bf w}\rangle \\ \\
{\bf v}^{\top}{\bf w} &=& ({M\bf v})^{\top}{M \bf w} \\ \\
{\bf v}^{\top}{\bf w} &=& {\bf v}^{\top}M^{\top}M{\bf w} \\ \\
{\bf v}^{\top}{\bf w} &=& {\bf v}^{\top}(M^{\top}M){\bf w}
\end{eqnarray*}
 This means that angles and distances are preserved by a linear transformation, with matrix representation $M$, if and only if, $M^{\top}M = E$, where $E$ is the identity matrix.
Notice that $M^{\top}M = E \implies \det(M^{\top}M) = \det(E) \implies \det(M)^2 = 1$. This means that $\det(M) = \pm 1$ and so all orthogonal matrices have determinant $-1$ or $+1$. The special orthogonal matrices, $\mathrm{SO}(n)$, have positive determinant and preserve orientation. 
A: The groups $SO(3)$ and $O(3)$ are different, in particular the first group is a subgroup of the second one of index $2$.
In particular
$$O(3)=\big\{\psi\in GL(3,\mathbb R):\psi^t\circ\psi=id \text{ and } \det\psi=\pm1\big\}$$
$$SO(3)=\big\{\psi\in GL(3,\mathbb R):\psi^t\circ\psi=id \text{ and } \det\psi=1\big\}$$
The geometric interpretation is the following thing: $O(3)$ is the group of the isometries of $2-$dimensional sphere, whereas $SO(3)$ is the subgroup of the orientation preserving isometries of the sphere.
A: $\psi^t \circ \psi=id$ entails that $\forall x \in \mathbb{R}^3, \|\psi(x)\|=\|x\|$. So $\psi$ is an isometry in $\mathbb{R}^3$.
For the second question, the answer is no. Example: $\phi=2  id$ is bijective and has only one fixed point. But $\phi^t \circ \phi=4id$.
Another example: $\phi=\left(\begin{matrix}2 &0 &0 \\ 0&2&0 \\ 0&0&1\end{matrix}\right)$,
$\mathbb{R}e_3$ is fixed. But $\phi^t \circ \phi=\left(\begin{matrix}4 &0 &0 \\ 0&4&0 \\ 0&0&1\end{matrix}\right) \neq id$
A: Since you have asked for a geometric interpretation, I am giving an intuitive/informal explanation that is usually given to physics students.
First consider the rotation matrix that describes a rotation by angle $\theta$ as seen from the $+z$ axis. It is not hard to see that this rotation can be represented as
$$R = \begin{pmatrix}
\cos\theta & -\sin\theta & 0\\
\sin\theta & \cos\theta & 0 \\
0 & 0 & 1\end{pmatrix}$$
Taking the transpose operation is equivalent to setting $\theta \rightarrow -\theta$. So one can think about $R^T$ as being a rotation by $-\theta$ as seen from the $+z$ axis.
With this picture in mind $R^T R = R R^T= \mathrm{id}$ follows.
Any arbitrary rotation can be described by a composition of rotations about the axes (see Euler angles). So say $\psi = R_x R_y R_z$. Then $\psi^T = R_z^TR_y^TR_x^T$. We have already seen that $R_z^TR_z=\mathrm{id}$; there will be similar identities for $x$ and $y$. Then $\psi^T\psi = \mathrm{id}$ follows. 
One can think of $\psi$ as being specified by a 3 dimensional vector (along the axis of rotation) and an angle of rotation. From our "proof", $\psi^T$ represents a rotation about the same axis but the opposite angle. This generalizes the observation made when the rotation was about the $z$ axis.
Edit: Wikipedia's page on Rotation matrix has some nice examples with pictures.
