I was reading the famous paper of Bhama Srinivasan "The characters of the finite symplectic group $Sp(4,q).$" An AMS link for the paper is here. $Sp(4,q)$ is the group of all invertible $4\times 4$ matrices $X$ over $F_q$ satisfying $XAX^t=A,$ where

$A = \left( \begin{array}{cccc} 0 & 1 & 0 & 0\\ -1 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & -1 & 0 \end{array} \right)$

On pages 489-491, the author mentions conjugacy classes of the group $Sp(4,q).$ In the second last row on page 490, the author mentions the representative of a conjugacy class denoted by $B_9(i)$ as the following matrix

$B_9(i)=\left( \begin{array}{cccc} \gamma^i & 0 & 1 & 0 \\ 0 & \gamma^{-i} & 0 & 1 \\ 0 & 0 & \gamma^i & 0\\ 0& 0& 0& \gamma^{-i} \end{array} \right)$

where $i\in \{1,2,\dots,\frac{1}{2}(q-3)\} $ and $\gamma \in F_q.$

Now if I compute $B_9(1)AB_9(1)^t,$ I get

$ \left( \begin{array}{cccc} 0 & 2& 0 & \gamma^{-1} \\ -2 & 0& -\gamma & 0 \\ 0 & \gamma & 0 & 1\\ -\gamma^{-1}&0 &-1 &0 \end{array} \right) \neq A$

As per definition of $Sp(4,q), B_9(1)AB_9(1)^t$ should be equal to $A.$ I am not getting what I am doing wrong. I will be grateful for any help in understanding my mistake or any other good reference for conjugacy classes of $Sp(4,q)$