Let $G=2^{2n}{:}Sp\_{2n}(2)$ be the split extension, where the symplectic group $Sp\_{2n}(2)$ acts naturally on the vector space $2^{2n}$. With the aid of GAP it turns out that the automorphism group $\textrm{Aut}(G)\cong G{:}2\cong 2^{2n+1}{:}Sp\_{2n}(2)=G\_1$ for $n=2, 3, 4$. Furthermore, the group $G\_1$ is not isomorphic to the affine subgroup $G\_2=2^{2n+1}{:}Sp\_{2n}$ of $Sp\_{2n+2}(2)$ which has a similar structure. Note that the group $G\_3=2^2{:}Sp\_2(2)\cong S\_4$ is complete and $G\_4=2^3{:}Sp\_2(2)\cong G\_3\times 2$, where $G\_4$ is the affine subgroup of $Sp\_4(2)$.


My question is how do one prove that $\textrm{Aut}(G)\cong G{:}2\cong 2^{2n+1}{:}Sp\_{2n}(2)$ for any $n\geq 2$ (if it is true) or is there already literature available on this topic.



The automorphism group of $2^{2n}{:}Sp_{2n}(2)$
Since ${\rm Sp}(2n,2)$ has trivial outer automorphism group and its natural module $M$ is absolutely irreducible, this follows from the fact that $|H^1({\rm Sp}(2n,2),M)| =2$.


You can find that result, for example, in Table 4.5 of


Cohomology of finite groups of Lie type, I,
Edward Cline; Brian Parshall; Leonard Scott,
Publications Mathématiques de l'IHÉS (1975)
Volume: 45, page 169-191
(see [here](https://eudml.org/doc/103939)),


but that might not be the earliest proof.


