Quaternion conjugation map is orthogonal linear transformation The following concerns an exercise from an undergraduate level textbook on Lie groups, which provides an elementary proof of $SU(2)/\mathbb{Z}_2 \cong SO(3)$. 
Think of $Sp(1)$ as the group of unit-length quaternions; that is, $Sp(1) = \left\{ q \in \mathbb H : |q| = 1 \right\}.$ For every $q \in Sp(1)$, show that the conjugation map $C_q: \mathbb H \to \mathbb H$, defined as $C_q(V) = q V \overline{q}$, is an orthogonal linear transformation. Thus, with respect to the natural basis $\left\{ 1, i, j, k \right\}$ of $\mathbb H$, $C_q$ can be regarded as an element of $O(4)$. 
My attempt so far has been as follows: To show it is an orthogonal linear transformation, we must show $\langle C_q(V), C_q(W) \rangle = \langle V, W \rangle$ for $V, W \in \mathbb H$. So we have 
$LHS = (qV \overline q) \cdot (\overline{qW \overline q}) = qV \overline q \cdot q \overline W \overline q = q V \overline W \overline q,$ 
and I just can't seem to figure out what to do from here. We know $q$, $\overline q$ are unit-length, but that does not mean they commute with $V, \overline W.$ I have a feeling that perhaps $V$ and $W$ must be pure imaginary quaternions, but even making that assumption doesn't seem to help. If anyone has any insight, I'd be very grateful. It's possible I'm using a bad/incorrect definition of orthogonal linear transformation, but it definitely seems like the error in question is something simple/obvious. Thanks for your time. 
 A: You have $\langle V,W \rangle = \frac{1}{2}(V \overline{W} + W \overline{V})$. Therefore
$$2\langle C_q(V),C_q(W) \rangle = q V\overline{q} q \overline{W} \overline{q} + q W \overline{q} q \overline{V} \overline{q} = q V\overline{W} \overline{q} + q W\overline{V} \overline{q} = q (V \overline{W} + W \overline{V}) \overline{q} \\ = q 2 \langle V,W \rangle \overline{q} .$$
But $2\langle V,W \rangle \in \mathbb{R}$, hence
$$q 2 \langle V,W \rangle \overline{q} = 2 \langle V,W \rangle q \overline{q} = 2 \langle V,W \rangle .$$
Added on request:
Write $V = a + bi + cj + dk, W = a' + b'i + c'j + d'k$. Then by definition $\langle V, W \rangle = aa' +bb' + cc' + dd' \in \mathbb{R}$. Moreover $V\overline{W} = (a + bi + cj + dk)(a' - b'i - c'j - d'k) = aa' +bb' + cc' + dd' + r$ with a suitable $r = ui + vj + wk$. Hence $\langle V, W \rangle = \text{Re}(V\overline{W})$. But $\text{Re}(V\overline{W}) = \frac{1}{2}(V\overline{W} + \overline{V\overline{W}}) = \frac{1}{2}(V\overline{W} + W\overline{V})$. 
