I have reason to believe that there is only one$(2^{2s+2},2^{2s+1}-2^s,2^{2s}-2^s)$- difference set (based on experimentation in GAP), up to equivalence/complementation, in any elementary 2-group of order $2^{2s+2}$. Note that these are so-called Hadamard difference sets. Is this indeed the case?
The extensive size of literature on difference sets in Abelian groups leads me to believe that, if this is true, it has already been proven. However, based on a short literature search, I have found nothing either confirming nor disproving my conjecture.
I have a rough idea about how to go about proving this claim, so if you are well-versed in difference set literature, and have seen nothing deciding this question in either direction, I am also very interested in knowing that, as I am mainly trying to decide if attacking this problem in detail is a worthwhile pursuit.
Definition: Let $G$ be a group of order $v$, and let $D \subset G$ be a subset of size $k$ such that the set $\{d_i d_j | d_i,d_j \in D\}$ contains each nonidentity element of $G$ $\lambda$ times. Then, we call $D$ a $(v,k,\lambda)$-difference set in $G$.
Example: If $G$ is the group $C_2^4$, and we think of its elements as binary $4$-tuples, then the set $D = \{(1,1,0,0),(1,1,1,0),(1,1,0,1),(0,0,1,1),(1,0,1,1),(0,1,1,1)\}$ is a $(16,6,2)$-difference set in $G$.
Definition: We say that two difference sets, $D_1,D_2 \subset G$, are equivalent if $D_1 = g D_2^\alpha$, where $g \in G$ and $\alpha \in Aut(G)$.
Example: $D^\prime = \{(0,0,0,0),(0,0,1,0),(0,0,0,1),(1,0,1,1),(0,1,1,1),(1,1,1,1)\}$ is a difference set in $C_2^4$, and it is equivalent to $D$ (above) because $D^\prime = (1,1,0,0)D$.
