I am interested in the finite unitary reflection group $G= G\_{32}$, the group No. 32 in Table VII on page 301 of the paper:
[Shephard, G. C.; Todd, J., A. Finite unitary reflection groups. Canad. J. Math. 6 (1954), 274–304](https://www.cambridge.org/core/journals/canadian-journal-of-mathematics/article/finite-unitary-reflection-groups/4A59731DC35C3FEC99F39EB2C009FE3E). 


This is a group of order $2^7 3^5 5 = 155520$. Its commutator subgroup $H=(G,G)$ is of index 3, and a computer calculation shows that $H$ is isomorphic to ${\rm Sp}(4,3):={\rm Sp}\_4({\Bbb F}\_3)$, the symplectic group of of $4\times 4$ matrices over the finite field ${\Bbb F}\_3$.


This group $G$ is given with a faithful 4-dimensional complex representation 
 $$\rho: G\to {\rm GL}(4, {\Bbb C}).$$
Moreover, it is is stable under the standard complex conjugation in ${\Bbb C}^4$, and so we obtain an involutive automorphism (an automorphism of order 2) $\ \sigma\colon H\to H$. 


I am trying to *guess* this involution $\sigma$ and to compute the first nonabelian cohomology set $H^1(\langle\sigma\rangle, H)$. A computer calculation shows that the $H^1$ is trivial, and I would like to understand this without computer.



> 
> **Question 1.** What are the nontrivial 4-dimensional complex representations 
>  of the finite group ${\rm Sp}(4,3)$?
> 
> 
> **Question 2.** What are the involutive automorphisms of ${\rm Sp}(4,3)$ ? In particular, is it true that all nontrivial involutive automorphisms of ${\rm Sp}(4,3)$ 
>  come from elements of order 2 in the projective symplectic group ${\rm PSp}\_4({\Bbb F}\_3)$ ?
> 
> 
> **Question 3.** Which of those involutive automorphisms of $H={\rm Sp}(4,3)$ can come from the complex conjugation in a 4-dimensional complex representation of $H$?
> 
> 
> 


Feel free to migrate this elementary question to Mathematics StackExchange.com.



Finite group ${\rm Sp}_4({\Bbb F}_3)$: involutions coming from a 4-dimensional complex representation
We can take $H={\rm Sp}(4,3)$ to be the group $\{ A \in {\rm GL}(4,3) \mid AFA^{\mathsf T} = F\}$, where $$F=\left(\begin{array}{rrrr}0&0&0&1\\0&0&1&0\\0&-1&0&0\\-1&0&0&0\end{array}\right)$$ is the matrix of the preserved symplectic form.


The the matrix $$C =\left(\begin{array}{rrrr}1&0&0&0\\0&1&0&0\\0&0&-1&0\\0&0&0&-1\end{array}\right),$$ satisfies $CFC^{\mathsf T}= -F$, it normalizes and induces an involutory outer automorphism of $H$, and $\langle H,C \rangle$ is the *conformal* symplectic group, which I prefer to denote by ${\rm CSp}(4,3)$ (although it is sometimes written as ${\rm GSp}(4,3)$).


There are two dual 4-dimensional complex representations of $H$, which are interchanged by the outer automorphism induced by $C$, so this appears to be the automorphism that you are looking for.


From your description, I think the only possible structure of the group $G$ is the direct product $H \times C\_3$.


To answer you specific questions, I am not sure what you are looking for in Question 1.


For Question 2, the full automorphism group group of $H$ is the image of ${\rm CSp}(4,3)$ mod scalars, which we can denote by ${\rm PCSp}(4,3)$: it has order $2|{\rm PSp}(4,3)| = 51840$. The involutory automorphism in question is an outer automorphism, and is not induced by an element of ${\rm PSp}(4,3)$.


For Question 3, I am not completely sure. There are actually two conjugacy classes of involutory outer automorphisms of $H$, one of which is induced by the matrix $C$ above, and I am not sure whether both can be induced by complex conjugation or only one of them.


An example of an element of ${\rm CSp}(4,3)$ that induces an involutory automorhism from the other class is $$C' =\left(\begin{array}{rrrr}0&0&1&0\\1&0&0&-1\\-1&0&0&0\\0&1&1&0\end{array}\right).$$ This has order 4 in ${\rm CSp}(4,3)$, but its square is $-I$, so it induces an involutory automorphism. It is interesting that its centralizer in $H$ has order $720$, whereas the centralizer of $C$ has order $48$. That might be useful in deciding which automorphism is induced by complex conjugation.


