= Solution
For <equal-length crystallographic root configurations in a plane>, taking $u=v$ in the closure condition gives $S_v(v)=-v$, so every vector occurs with its negative. For fixed $v\in\Phi$, a vector $u\in\Phi$ has $\langle u,v\rangle\in\{-1,-1/2,0,1/2,1\}$, because both vectors are <unit vectors>. Thus its angle relative to $v$ belongs to the finite list $0,\pm\pi/3,\pm\pi/2,\pm2\pi/3,\pi$.
If all vectors are collinear, the set is exactly $\{v,-v\}$, giving two opposite points of the <unit circle>.
If an <orthogonal> vector $w$ occurs, closure supplies $\pm v,\pm w$. A <unit vector> making an angle of $\pi/3$ or $2\pi/3$ with $v$ would have <inner product> $\pm\sqrt3/2$ with $w$, contradicting integrality of twice that <inner product>. Thus these four points are the whole set: the vertices of a <square>.
In the remaining case a noncollinear pair makes an angle of $\pi/3$ or $2\pi/3$. Changing a sign, take their angles as $0,\pi/3$. Reflecting the second in the line perpendicular to the first supplies the angle $2\pi/3$, and negation supplies the other three hexagon vertices. No perpendicular vector can occur, by the preceding inner-product obstruction. The initial finite list now leaves only these six points. Conversely the opposite pair, <square> and <regular hexagon> each satisfy both <reflection> closure and the integrality condition. Hence
$$
\boxed{|\Phi|\in\{2,4,6\}.}
$$
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