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 $L = \mathbb C$.]