M(a b c) =
!a 
   !b
      !c
 a  b
 a     c

best assignment: 4/5 Jmax

Apply full permutation group of 3 elements to M
and concatenate all formulas:

Sym(M) =
M(a b c) identity
M(a c b) transpose b c
M(c b a) transpose a c
M(b a c) transpose a b
M(c a b) rotate right
M(b c a) rotate right

Sym(M) is symmetric.

Why has Sym(M) a lower best satisfaction ratio
than M.
Because for the best assignment Jmax for M:
fsat(M,Jmax) >= fsat(Sym(M),Jmax)
Proof: for each of the permutations of M, the satisfaction ratio
is <= fsat(M,Jmax)

Symmetrization (go from M to Sym(M))
lowers or keeps the same the best satisfaction ratio.
It stays the same if M is already symmetric.