YES

2 decompositions
 #0 -----------
   4: a() -> b()

 #1 -----------
   1: f(b()) -> c()
   2: c() -> c()
   3: f(c()) -> b()

@Mutually Orthogonal
--- R
   4: a() -> b()

--- S
   4: a() -> b()


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

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