Solution: Let \[ A = \begin{bmatrix} a + \iota b & c + \iota d \\ -c + \iota d & a - \iota b \\ \end{bmatrix} \] be a unit in $H$. Therefore, $A^{-1} \in H$. Since $A^{-1} $ exists, the determinant of $A$ must be nonzero, that is, $a^2 + b^2 + c^2 + d^2 \neq 0$. Note that \[ A^{-1} = \frac{1}{a^2 + b^2 + c^2 + d^2} \begin{bmatrix} a - \iota b & -c - \iota d \\ c - \iota d & a + \iota b \\ \end{bmatrix} \in H. \]

So $A^{-1} \in H \iff a^2 + b^2 + c^2 + d^2 \in \mathbb{Z} $. Since $a,b,c,d\in \mathbb{Z} $, the following are the possibilities. \begin{align*} a = \pm 1,\ b = c = d = 0 \\ b = \pm 1,\ a = c = d = 0 \\ c = \pm 1,\ a = b = d = 0 \\ d = \pm 1,\ a = b = c = 0. \end{align*} Therefore, the units in $H$ are \begin{align*} \begin{bmatrix} \pm 1 & 0 \\ 0 & \pm 1 \\ \end{bmatrix}, \begin{bmatrix} \pm \iota & 0 \\ 0 & \mp \iota \\ \end{bmatrix}, \begin{bmatrix} 0 & \pm 1 \\ \mp 1 & 0 \\ \end{bmatrix}, \begin{bmatrix} 0 & \pm \iota \\ \pm \iota & 0\\ \end{bmatrix}. \end{align*}