Regarding the Galois group of the factor $g(x)=g(q,x)$of these Fekete polynomials that is conjectured to be irreducible, here is PARI code that counts, for each prime $q \equiv 1 \bmod 4$, $17 \leq q \leq 41$$17 \leq q \leq 100$, the number of primes $p$ for which $g(q,x)$ modulo $p$, has $n$ linear factors, for primes $p$ up to half a million, stored in the matrix $c[q,n]$$c[q,n+1]$, done for $2\leq n\leq 12$$0\leq n\leq 14$.
f(m,x)=sum(i=1,m-1,kronecker(i,m)*x^i))
g(m,x)=f(m,x)/(x*(x-1)^2*(x+1))
for(n=2n=0,1214,forprime(q=17,41100,if(q%4==1,forprime(p=2,500000,if(matsize(polrootsmod(g(q,x),p))==[n,1], c[q,n]++n+1]++)))))
If I did not make any mistakes, then, for these fournine primes $q=17,29,37,41$$q=17,29,37,41,53,61,73,89,97$, here is the $4$$9$ by $12$$15$ matrix showing $n=1$$n=0$ to $12$$14$.
[0[25204 3 12634 0 3084 0 538 0 72 0 3 0 0 0 0]
[0[25360 2 12486 0 3101 0 530 0 50 0 9 0 0 0 0]
[0[25133 1 12637 0 3187 0 519 0 59 0 2 0 0 0 0]
[0[25320 2 12519 1 3081 0 543 0 65 0 6 0 1]1 0 0]
What appears to happen is that $n=0$ which is not shown here and which I only tried for $q=17$ gives about 60%. Please let me know if you find any mistakes, thanks![25293 2 12449 0 3206 0 500 0 81 0 6 0 1 0 0]
[25176 2 12603 0 3157 0 527 0 69 0 4 0 0 0 0]
[25110 3 12564 0 3274 0 516 0 65 0 5 0 1 0 0]
[25286 3 12554 0 3116 0 502 0 71 0 6 0 0 0 0]
[25112 3 12711 0 3132 0 526 0 49 0 5 0 0 0 0]