The real parts and the imaginary parts of two quaternions Let $x=x_0+ix_1+jx_2+kx_3$ and $y=y_0+iy_1+jy_2+ky_3$ be qauternions. The real parts of $x,y$ are $y_0$, $x_0$, respectively. Also, we denote $\Re(x)$, $\Re(y)$, respectively. The imaginary parts of $x,y$ are $ix_1+jx_2+kx_3$, $iy_1+jy_2+ky_3$, respectively. Also, we denote $\Im(x)$, $\Im(y)$.
We want to show that
$y$ is similar to $x$ i.e., $y=axa^{-1}$ for some $a\neq 0$ quaternion if and only if $\Re(x)=\Re(y)$ and $|\Im(x)|=|\Im(y)|$.
 A: First of all, notice that $axa^{-1}=\frac{a}{|a|}x\frac{a^{-1}}{\frac{1}{|a|}}=\frac{a}{|a|}x\frac{a^{-1}}{|a^{-1}|}$
, so without loss of generality, we can suppose $a$ is a unit length quaternion. With this assumption, $a^{-1}=\bar a$.
Then $\Re (axa^{-1})=\frac{axa^{-1}+\overline{axa^{-1}}}{2}=\frac{axa^{-1}+a\overline{x}a^{-1}}{2}=a(\frac{x+\bar x}{2})a^{-1}=a(\Re(x))a^{-1}=\Re (x)$.
As a direct consequence, any conjugate of a pure quaternion is again a pure quaternion. Furthermore, conjugation preserves the length of the quaternion:
$|axa^{-1}|^2=axa^{-1}\overline{axa^{-1}}=axa^{-1}a\bar xa^{-1}=ax\bar xa^{-1}=a|x|^2a^{-1}=|x|^2$, and taking the square root of both sides confirms this. So, conjugation fixes the real part and moves the pure quaternion part to another pure quaternion of the same length.
Conversely, if $x$ and $y$ are quaternions with identical real part and whose pure quaternion parts have the same length, you can find a (unit) quaternion $a$ such that $y=axa^{-1}$.
You simply think of the pure quaternion parts as lying in $3$-dimensional real space, and find a rotation (there are infinitely many solutions) in terms of a quaternion $a$ moving the pure quaternion part of $x$ onto the pure quaternion part of $y$. Conjugation with this $a$ automatically preserves the real part and moves the pure quaternion part so that $y=axa^{-1}$ holds.
