Real matrices that lie in the image of the inclusion homomorphism $\rho_n: M_n(\mathbb C) \to M_{2n}(\mathbb R)$ are called complex linear real matrices. It is easy to see that a real matrix is complex linear if and only if it commutes with $I = \rho_n(iI)$.
In analogy to this I am now studying the quaternionic inclusion $M_n(\mathbb H) \to M_{4n}(\mathbb R)$ using the inclusion $\psi_n: M_n(\mathbb H) \to M_{2n}(\mathbb C)$. If $i,j,k$ denote the unit quaternions then I want to find matrices $I$ and $J$ such that a real matrix is quaternionic linear if and only if it commutes with $I$ and $J$.
In $1$ dimension I tried $I=\rho_{2n} \circ \psi_n (iI)$ and $J=\rho_{2n} \circ \psi_n (jI)$ but the problem then is that $I^2 \neq -1$ and $J^2 \neq -1$.
Why does $I=\rho_{2n} \circ \psi_n (iI), J=\rho_{2n} \circ \psi_n (jI)$ not work in the quaternionic case? Is there an insightful geometric (or other) explanation? For the inclusion of complex matrices into real matrices setting $J=\rho_{2n} (iI)$ worked.
Edit For a definition of $\rho_n$:
define $\rho_n : M^n(\mathbb C) \to M^{2n}(\mathbb R)$ as $A_{ij}\mapsto \begin{array}{cc} a_{ij} & b_{ij} \\ -b_{ij} & a_{ij} \end{array}$ if $A_{ij}=(a_{ij} + i b_{ij})$
and $\color{blue}{\psi_n}: M^n(\mathbb H) \to M^{2n}(\mathbb C)$ as $A_{ij}\mapsto \begin{array}{cc} a_{ij} & b_{ij} \\ -\overline{b_{ij}} & \overline{a_{ij}} \end{array}$ if $A_{ij}=(a_{ij} + b_{ij}j)$
Edit 2 (in response to the anwer)
Let $$I = \rho(\color{blue}{\psi(i)})= \rho\left ( \begin{array}{cc} \color{blue}{i} & \color{blue}{0} \\ \color{blue}{0} & \color{blue}{-i} \end{array}\right)$$ Then $$ I = \left ( \begin{array}{cccc} 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0\\ 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 \end{array} \right )$$
so that
$$ I^2 = -1$$
Edit 3
After the discussion with Incnis Mrsi I calculated $J^2=I^2 =-1$ for $J=\rho_{2}(\psi_1(j))$ and $I=\rho_{2}(\psi_1(i))$. I am confused that this seems to work. The reason why I asked this quetion is the following passage in Tapp's matrix groups for undergraduates:
In particular,
why is $I\neq\rho_{2n}(\psi_n(i))$ and $J\neq\rho_{2n}(\psi_n(j))$ for $n>1$? (for $n=1$, apparently it works as I just verified).