How strong are nonary numbers?
Python 296
Not actually too inefficient, it only checks as many digits as it needs to.
n=raw_input();s=int(len(n)**.5);t=0
for i in range(s**2):
    l=[]
    while i not in l:l.append(i);c=n[i];i=(i%s+(-1 if c in'456'else 1 if c in'218'else 0))%s+((i/s+(-1 if c in'432'else 1 if c in'678'else 0))%s)*s
    t=t*9+len(l)-l.index(i)
print'EGL'[cmp(t,long(n,9))]

As for numbers equal to their strength, I think the only solutions are, for every N x N square up to N=8 a square containing N in every space. My thinking is that since every number in a loop must be the same number (the length of the loop) each loop would have to be all in one direction. This of course means that the loop's size must be N (and each element must be N). I'm pretty sure that this logic can be applied to squares and loops of any size, meaning there are no squares equal to their strength other than the first 8.