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.

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

Reference

[1] Peter Kleidman, Martin Liebeck, The subgroup structure of the finite classical groups London Mathematical Society Lecture Note Series, 129. Cambridge: Cambridge University Press, pp. x+303 (1990), ISBN:0-521-35949-X, MR1057341, Zbl 0697.20004.