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Does there exist a quaternion $q$ on the unit sphere such that, given the vanilla complex plane $\mathbb{C}$, $q\mathbb{C}q^{-1} = \bar{\mathbb{C}}$?

Motivation: ordinarily, the plane is rotated by multiplying it by a complex number in the unit circle. Taking the conjugate amounts to flipping the plane along the real axis, which can be viewed as a rotation.

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This looks similar to the definition of a normal subgroup. –  PEV Mar 18 '11 at 20:16
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up vote 2 down vote accepted

As a variation on Joseph's answer:

Since $j \cdot 1 \cdot j^{-1} = 1$ while $j \cdot i \cdot j^{-1} = -i,$ we see that $j z j^{-1} = \overline{z}$ for any complex number $z$.

[This is a special case of a more general theorem, by the way: if $A$ is a central simple algebra over a field $K$, and $L$ is a field extension of $K$ contained in $A$, then any field automorphism of $L$ over $K$ can be realized via conjugation by an invertible element of $A$. This particular case is $A = \mathbb H$, $K = \mathbb R$, and $L = \mathbb C$.]

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I don't know if there's a single rotation that gives you what you want, but given any complex number $z$, the complex conjugate of $z$ is given by

$$\bar{z}=-\frac{1}{2}(z-izi^{-1}-jzj^{-1}-kzk^{-1}).$$

Complex conjugation is a linear combination of quaternion rotations, at least. Note that this equation actually works for all quaternions.

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