Let $p$ be an odd prime. What's the condition on $q$ for $p^{1+2r}.\operatorname{Sp}(2r,p)\leqslant \operatorname{GU}(p^r,q)$. I did some computation and seemed that $q\equiv -1$(mod $p$) does give the embedding. I feel that there is work out there about it? Or it is an obvious question and I'm being silly. I did check Kleidman and Liebeck's book on maximal subgroups. It just hasn't provided much help. Thank you.