YES

1 decompositions
 #0 -----------
   1: f(x) -> g(x)
   2: f(x) -> h(f(x))
   3: h(f(x)) -> h(g(x))
   4: g(x) -> h(g(x))

@Strongly Commuting
--- R
   1: f(x) -> g(x)
   2: f(x) -> h(f(x))
   3: h(f(x)) -> h(g(x))
   4: g(x) -> h(g(x))

--- S
   1: f(x) -> g(x)
   2: f(x) -> h(f(x))
   3: h(f(x)) -> h(g(x))
   4: g(x) -> h(g(x))