$USp(N)$ with $N$ an even integer is defined as the group of unitary matrices $M$ that satisfy $M^TJM=J$, where $M^T$ is the transpose of $M$ and $J$ is the anti-symmetric $N$-by-$N$ matrix \begin{equation} J= \left( \begin{array}{cc} &I\\ -I& \end{array} \right) \end{equation} where $I$ is the $\frac{N}{2}$-by-$\frac{N}{2}$ identity matrix.
$USp(N)$ group has $\frac{N(N+1)}{2}$ generators. For $USp(4)$, the generators can be represented as $\Gamma_{ab}$ which are defined through another set of matrices $\Gamma_a$ by \begin{equation} \Gamma_{ab}=\frac{1}{2i}(\Gamma_a\Gamma_b-\Gamma_b\Gamma_a) \end{equation} and $\Gamma_a$ themselves satisfy $\Gamma_a\Gamma_b+\Gamma_b\Gamma_a=2\delta_{ab}$.
In this case we can verify that $J\Gamma_{ab}J=\Gamma_{ab}^T$ and $J\Gamma_{a}J=-\Gamma_{a}^T$. Moreover, $\sum_{ab}\Gamma_{ab}\Gamma_{ab}$ and $\sum_{a}\Gamma_{a}\Gamma_{a}$ are invariant under the fundamental representation of $USp(4)$, namely, $\sum_{ab}\Gamma_{ab}\Gamma_{ab}=U^\dagger\sum_{ab}\Gamma_{ab}\Gamma_{ab}U$ and $\sum_{a}\Gamma_{a}\Gamma_{a}=U^\dagger\sum_{a}\Gamma_{a}\Gamma_{a}U$ for any $USp(4)$ matrix $U$.
My question is: for any even $N>2$, can we always find some Hermitian matrices $\Gamma_a$ so that $J\Gamma_aJ=-\Gamma_a^T$ and $\sum_a\Gamma_a\Gamma_a$ is invariant under the fundamental representation of $USp(N)$ in the above sense?
My conjecture to this question is that there are indeed such $\Gamma_a$, and more precisely, there are $N+1$ of them and they can still be used to derive the $\frac{N(N+1)}{2}$ generators of $USp(N)$ by $\Gamma_{ab}\propto\Gamma_a\Gamma_b-\Gamma_b\Gamma_a$.
I appreciate if any one will check this conjecture and answer this question.