Start with the the weyl group Z/2 xGL(3).
1) there is only one mod 2 lattice
2) all reflections are constrained
3) therefore all calculations of centers and components of centralizers go
   as in the std case of the example we constructed.
4) Could calculate that rk 1 centralizer has weyl group of Spin(7), but we
don't know yet that Spin(7) is determined by its weyl group + info on ker of
mod 2 reduction map ( I think this is do-able however).
5) Plan 2: work with rank 2 centralizer. Weyl group is z/2xz/2xz/2, so
conclude
is a form of (SU(2)^3/z/2z . Which one? No reflection can reduce to zero, so
only choice is the diagonal central z/2z. By previous work this is determined
up to
homotopy equivalence.
6) Now look at the normalizer of the rk 2 elementary. This Z/2Z x [[GL_2 , *
],[0,1]],
where * is arbitrary length 2 vector to fill in last column of 3x3 matrix.
This gives us the normalizer in a sequence
1 --> H --> N_X(E) --> GL_2 = \Sigma_3 --> 1

where I believe that GL_2 permutes the SU(2) factors in H.
Now euler characteristics tell you that H^*BX --> H^*BN_X(E) is monic.
I f the above extension is split, then the 2 rank of N_X is only 4. and
likewise
for X. However, I believe that there is also a nonsplit extension of the
above
also, and I don't see how to decide. You really just need it over a
transposition,
but I seem to get an obstruction in H^3(BZ/2, Z/2xZ/2).

 Whether it's split or nor, one can calculate from the centralizer
 of  a rank 3
elem. abel that there is rk 4 elem. ab. that is  selfcentralizing .

If the N_X is split, you could argue that its cohomology is detected on elem.
ab.
subgroups, and therefore same for X. 

Not much to show for 550 miles of driving.
Clarence
uBut to finish up, if it is split, then one can go ahead and use that there
is a
self-centralizing (z/2z)^4, calculating from the weylgroup.