Symplectic group over finite field is defined as group preserving non-degenerate antisymmetric bilinear form on $\mathbb F_q^{2n}$. How could we define this group using quaternions ? This should be group of $n\times n$ matrices with quaternion entries preserving hermitian form.
I know that there might be troubles with definition of quaternions over finite field. One approach is.to use Cayley-Dickson formula. Second is to use some 4-dimensional subalgebra of octonions.
Here are some hints for $q=2$.
Let us define quaternions $\mathbb H_2$ as algebra $M_2\mathbb F_2$. It is generated by two letters $t,u$ with conjugation and relations $\bar t=1+t, \bar u =u, ut=\bar tu$. In this case I claim that group of matrices $\{A A^*=I\}$ form $S_{2n}(2)$ finite simple group. $A^*$ denote transposed comjugated matrix to $A$.
Complex case
Let $\mathbb C_1$ be subalgebra of $\mathbb H_2$ generated by zero divisor $a$ such that $aa=a$. In this case $\bar a=1+a$ is second zero divisor in algebra. Subgroup of $\{A A^*=I\}$ with elements from $\mathbb C_1$ form $L_n(2)$ finite simple group.
Let $\mathbb C_2=\mathbb F_4$ be subalgebra of $\mathbb H_2$ generated by $t$. Subgroup of $\{A A^*=I\}$ with elements from $\mathbb C_2$ form $U_n(2)$ finite simple group.
Geometry
Let us define projective space over quaternions as $\mathbb HP_q^n=\frac{S_{2n}(2)}{S_{2n-2}(2)\times S_2(2)}$
The size is $20, 336, 5440$ for $n=1,2,3$. General formula is $2^{2n}\frac{2^{2n-2}}{3}$.