A matrix in $SO(3)$ is uniquely determined by what it does to $4$ rays The group $\mathrm{SO}(3)$ acts smoothly on the unit sphere $S^2\subset \mathbb{R}^3$. If we consider the involution $\sigma$ on $S^2$ that sends a point to the diametrically opposite point, then $S^2/\sigma$ is diffeomorphic to $\mathbb{R}P^2$.
Is an element of $\mathrm{SO}(3)$ uniquely determined by what it does to the $4$ points of $S^2/\sigma$ corresponding to $\frac{1}{\sqrt{3}}(1, 1, 1)$, $\frac{1}{\sqrt{3}}(1, 1, -1)$, $\frac{1}{\sqrt{3}}(1, -1, 1)$, $\frac{1}{\sqrt{3}}(1, -1, -1)$? If we did not take the quotient by $\sigma$, the answer would be positive because the vectors in question span $\mathbb{R}^3$.
 A: The vectors more than span $\Bbb R^3$. Because there are 4 of them, they are not linearly independent and the action on 3 of them will determine the action on the 4th. But that is true for any operator. $SO(3)$ satisfies additional restraints that can be used to find the operator with even less information. 
Start by examining what happens for $\Bbb R^3$. $SO(3)$ is the set of rotations, and all rotations are about some axis. All rotated vectors do not change their angle to this axis, therefore given that vector $a$ is rotated to vector $b$ (so $\|a\| = \|b\|$), the axis of rotation must make the same angle with both. The set of all such axes lies in a plane that passes through their sum $a + b$, and their cross product $a \times b$. A normal to that plane is given by $$(a + b) \times (a \times b) = (b\cdot(b - a))a + (a\cdot(a -b))b$$
Any line through the origin and perpendicular to this vector is the axis of a rotation carrying $a$ to $b$. Given which axis of rotation was used, $a$ and $b$ determine the angle and direction of the rotation.
So a single vector will not be enough to fully determine the rotation. Suppose we also know that $c$ rotates to $d$. Once again, this determines another plane whose lines through the origin can also be an axis of rotation. Assuming the plane is not the same as for $a$ and $b$, the two planes will intersect in a line, which must be the axis of rotation. Having identified the axis, you can write $a = a_\| + a_\perp, b = b_\| + b_\perp$ where $a_\|, b_\|$ are parallel to the axis and $a_\perp, b_\perp$ are perpendicular. Then $\frac {\|a_\perp \times b_\perp\|}{\|a_\perp\|\|b_\perp\|}$ will be the sine of the rotation angle. Of course $c$ and $d$ must determine the same rotation angle, or else no such rotation exists.
So - except for degenerate cases - the fate of two vectors under the operator is enough to specify which element of $SO(3)$ it is.
So what happens for $\Bbb {R}P^2$? The only difference is that instead of knowing that $a \mapsto b, c \mapsto d$, you know that $\pm a \mapsto \pm b, \pm c \mapsto \pm d$. So either $a$ rotates to $b$ or it rotates to $-b$. And either $c$ rotates to $d$ or it rotates to $-d$. This gives you 4 possible candidates for the rotation operator. $$(a\mapsto b, c\mapsto d)\\(a\mapsto b, c\mapsto -d)\\(a\mapsto -b, c\mapsto d)\\(a\mapsto -b, c\mapsto -d)$$
But again, except when $a_\perp \perp b\perp$, the angles that $a_\perp$ makes with $b_\perp$ and $-b_\perp$ will not be the same. And similarly for $c_\perp$ and $d_\perp$. Of the 4 cases, two will attempt to match angle for $a_\perp$ to $\pm b_\perp$ with the other angle for $c_\perp$ to $\pm d_\perp$, which will have no solution.
So given the fate of 2 points of $\Bbb RP^2$ under the action of an operator in $SO(3)$, except in degenerate cases, you can determine which operator up to being one of two choices.
The fate of a 3rd point, barring degeneracy, would be more than enough to tell which of the two operators was used.
