= Solution
For unit $u$, put $A=I-2uu^T$. Then $A^T=A$ and $A^2=I$, so $A$ is an <orthogonal matrix>. The affine <reflection in a hyperplane> is $R(x)=Ax+2cu$. Hence $R(x)-R(y)=A(x-y)$ preserves the <Euclidean norm> and distance, and $R(x)=x$ whenever $u\cdot x=c$. Thus it is an <Euclidean isometry> fixing the hyperplane pointwise.
For $p\ne q$, any reflection exchanging them must have normal parallel to $p-q$, and its fixed hyperplane must contain their midpoint. These requirements determine the perpendicular bisector uniquely:
$$
\boxed{u=\frac{p-q}{\|p-q\|},\qquad c=u\cdot\frac{p+q}{2},\qquad R(p)=q.}
$$
Changing both signs of $u,c$ leaves the same reflection. It belongs to the <orthogonal group> $O(n)$ exactly when its translation term vanishes, that is, $c=0$. Since $c=(\|p\|^2-\|q\|^2)/(2\|p-q\|)$,
$$
\boxed{R\in O(n)\iff\|p\|=\|q\|.}
$$
To prove the <finite reflection decomposition of a Euclidean isometry>, first note that a distance-preserving map $g$ fixing zero preserves inner products by the <polarization identity>. Its values on an orthonormal basis $e_i$ form an orthonormal basis, and $g(x)\cdot g(e_i)=x_i$ shows $g(x)=\sum_i x_i g(e_i)$. Thus $g$ is orthogonal.
Every <orthogonal transformation> is a product of at most $n$ linear reflections: if $g(e_1)\ne e_1$, reflect $g(e_1)$ to $e_1$ using the hyperplane normal $g(e_1)-e_1$, which passes through zero. The resulting orthogonal transformation fixes $e_1$, and its restriction to $e_1^\perp$ is handled inductively in dimension $n-1$. If $g(e_1)=e_1$, no initial reflection is necessary. Reflections on the complement extend by fixing $e_1$; the induction starts in dimension zero.
For an arbitrary isometry $f$, if $f(0)\ne0$, first reflect $f(0)$ to zero across its perpendicular bisector with zero. Composing this one affine reflection with $f$ gives an orthogonal transformation. If $f(0)=0$, skip the initial reflection. Therefore
$$
\boxed{f\text{ is a product of at most }n+1\text{ hyperplane reflections}.}
$$
The <glide reflection> $(x,y)\mapsto(x+1,-y)$ cannot use fewer than three: its determinant sign is negative, excluding zero or two reflections, while absence of fixed points excludes a single reflection. Three suffice by reflecting successively in $x=0$, $x=1/2$, and $y=0$.
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