YES

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

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

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