My computation shows:
2 and 3 are primitive root modulo 5;
Starting from 2, searching $2+5k$ that $(2+5k)^4 \not\equiv 1(mod 25)$ I get 20 roots: $2, 12, 17, 22, 27, 37, 42, 47, 52, 62, 67, 72, 77, 87, 92, 97, 102, 112, 117, 122$, i.e. omitting $7,32,57,82,107$.
Starting from $3$, excluding $18,43,68,93,118$ I get another 20 roots: $3,8,13,23,28,33,38,48,53,58,63,73,78,83,88,98,103,108,113,123$.
So there are totally 40 primitive roots modulo $125$, am i right?