YES

1 decompositions
 #0 -----------
   1: f(a()) -> f(f(a()))
   2: a() -> b()
   3: f(x) -> f(b())

@Strongly Commuting
--- R
   1: f(a()) -> f(f(a()))
   2: a() -> b()
   3: f(x) -> f(b())

--- S
   1: f(a()) -> f(f(a()))
   2: a() -> b()
   3: f(x) -> f(b())