Set of possible reflections of a vector.

Yep! Every reflection of $b$ has the same norm, as $H$ is orthogonal. Specifically, begin{align*} H^top H &= left(I - 2frac{vv^top}{|v|^2}right)^top left(I - 2frac{vv^top}{|v|^2}right) \ &= left(I - 2frac{vv^top}{|v|^2}right) left(I - 2frac{vv^top}{|v|^2}right) \ &= I - 4frac{vv^top}{|v|^2} + 4frac{v(v^top v) v^top}{|v|^4} \ &= I - 4frac{vv^top}{|v|^2} + 4frac{v v^top}{|v|^2} \ &= I = HH^top. end{align*} Therefore $|Hb| = |b|$ for all such matrices $H$.

On the other hand, choose any $c$ such that $|c| = |b|$, and let $v = b - c$. I claim that the $H$ corresponding to $v$ maps $b$ to $c$. We have begin{align*} Hb - c &= left(I - 2frac{vv^top}{|v|^2}right)b - c \ &= b - 2frac{vv^top b}{|v|^2} - c \ &= b - c - 2(v cdot b) frac{v}{|v|^2} \ &= b - c - 2((b - c) cdot b) frac{(b - c)}{|b - c|^2} \ &= ((b - c) cdot (b - c) - 2(b - c) cdot b)frac{b - c}{|b - c|^2} \ &= ((b - c) cdot (-b - c))frac{b - c}{|b - c|^2} \ &= (c cdot c - b cdot b)frac{b - c}{|b - c|^2} \ &= (|c|^2 - |b|^2)frac{b - c}{|b - c|^2} \ &= 0. end{align*}