DISCLAIMER: This is not homework.
I did this exercise here and need someone to check if my work is correct:
Is it possible to find a matrix $J\in M_{2n}(\mathbb C)$ such that the following diagram commutes?
$$\begin{array} \mathbb \mathbb H^n & \stackrel{g_n}{\longrightarrow} & \mathbb C^{2n} \\ \downarrow{ j} & & \downarrow{R_J} \\ \mathbb H^n & \stackrel{g_n}{\longrightarrow} & \mathbb C^{2n} \end{array} $$
from this book. The down arrow $j$ is multiplication by the unit quaternion $j$. Since the books convention is to use left scalar multiplication this is $jX$ for $X\in \mathbb H^n$. The map $R_J$ on the other hand is right multiplication by $J$: $R_J(X) = XJ$.
Let's assume that the map $g_n$ is $(a_1 + b_1j, a_2 + b_2 j, \dots) \mapsto (a_1, a_2, \dots, b_1, b_2, \dots)$.
My work:
I let
$J_{2n} = \rho_n (jI) = \left ( \begin{array}{cc} 0 & I \\ -I & 0 \end{array}\right ) $ where $\rho_n$ is the inclusion homomorphism $M_n(\mathbb H) \hookrightarrow M_{2n}(\mathbb C)$. Then
$$g_n(j(X + Yj)) = g_n(\overline{X}j + \overline{Y}j^2 = (-\overline{Y}, \overline{X})$$ and
$$ R_J(g_n(X + Yj)) = R_J((X,Y)) = (-Y,X)$$
so the diagram for this $J$ does not commute. (Note that if left multiplication by $j$ is replaced by right multiplication then the diagram commutes. )
I want to argue that this diagram commutes for no $J$. But how do I do this? I tried to let $J$ be arbitrary and consider the $n$-vector consisting of $i$. I denote it $i \in \mathbb C^n$ and I denote the (left) multiplication from the diagram by $L_j$. I think that
$$ L_j(i) = -k$$ and
$$ g_n^{-1}\circ R_J \circ g_n (i) = k$$
but even though $g_n (i) =i$ is clear to me I am stuck showing $R_J(i)= k$.
Note that your $g_n$ is invertible, so if such a $J$ exists then it should satisfy $$R_J(x)=(g_n\circ j\circ g_n^{-1})(x),$$ for every $x\in\Bbb{C}^{2n}$. Writing $x=(a_1,\ldots,a_n,b_1,\ldots,b_n)$, we can simply write out what the right-hand side is. As this is homework, I'll leave omit the tedious details, but you'll find $$(g_n\circ j\circ g_n^{-1})(x)=(-\overline{b_1},\ldots,-\overline{b_n},\overline{a_1},\ldots,\overline{a_n}),$$ where $\overline{z}$ denotes the complex conjugate of $z\in\Bbb{C}$. This is the composition of the map $$\Bbb{C}^{2n}\ \longrightarrow\ \Bbb{C}^{2n}:\ (a_1,\ldots,a_n,b_1,\ldots,b_n)\ \longmapsto\ (-b_1,\ldots,-b_n,a_1,\ldots,a_n),$$ which is linear and invertible, with coordinatewise complex conjugation, which is not linear (over $\Bbb{C}$!). It follows that $R_J$ is not linear, a contradiction, meaning that no matrix $J\in M_{2n}(\Bbb{C})$ making the diagram commute exists.