It known the Suzuki group $Sz(q)$, where $q=2^{2n+1}$ is of order $q^2(q^{2}+1)(q-1)$. By $2^{2}$ $=-1$ mod $5$, then $2^{2n}$ $=(-1)^{n}$ mod $5$. So $q=2^{2n+1}$ $=2(-1)^{n}$ mod $5$ and so $q^{2}+1=0$ mod $5$. Therefore $5$ always divide order of $Sz(q)$.

My question is: If $P$ is a Sylow $5$-subgroup of the group $Sz(q)$, then order of normalizer of $P$ in $Sz(q)$? Thanks.