TITLE: Geometric interpretation of the exceptional isomorphism $PSp(4,3)=PSU(4,2^2)$
QUESTION [15 upvotes]: It is well-known that there is an isomorphism between $PSp(4,3)$ (the symplectic group of dimension $4$ over $\mathbb F_3$) and  $PSU(4,2^2)$ (the unitary group defined by $4\times4$ unitary matrices over $\mathbb F_4$).
Question: Can the exceptional isomorphism be interpreted in (finite) geometry?
Known examples of geometric interpretation: 
The isomorphism $PSL(3,2)=PSL(2,7)$ is presented here.
The isomorphism $S_6=PSp(4,2)$ is explained in John Baez's article.
The isomorphism $A_6=PSL(2,9)$ can be explained by the Taylor graph of $PSL(2,9)$ acting on 10 points (i.e. the projective line of $\mathbb F_9$): this is a distance-regular graph on 20 vertices with intersection array $\{9,4,1;1,4,9\}$. Observing that this is also the intersection array of $J(6,3)$ and Johnson graphs are unique, it lets $S_6$ acts on the graph.
The isomorphism $PSU(3,3)=G_2(2)'$ can be explained by the unitary nonisotropics graph of $PΓU(3,q)$: for $q=3$, the intersection array is $\{6,4,4;1,1,3\}$, which is exactly that of a generalized hexagon of order 2. Furthermore, $GH(2,2)$ is unique up to duality, so it's also the Cayley generalized hexagon where $G_2(2)$ acts.

REPLY [13 votes]: I believe that a finite geometric proof is given by Jean Dieudonné here:
Dieudonné, Jean, Les isomorphismes exceptionnels entre les groupes classiques finis, Can. J. Math. 6, 305-315 (1954). ZBL0055.01904.
If you go to Section 7 of the given paper you will see the proof. Dieudonné displays the required isomorphism using a 1-1 correspondence between the 40 lines in $(\mathbb{F}_3)^4$, and the non-isotropic lines in $(\mathbb{F}_4)^4$ (non-isotropic with respect to a non-degenerate hermitian form).
A google-books version is here.