YES

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

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

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