YES
We show the termination of the TRS R:

  f(g(x)) -> g(g(f(x)))
  f(g(x)) -> g(g(g(x)))


-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: f#(g(x)) -> f#(x)

and R consists of:

r1: f(g(x)) -> g(g(f(x)))
r2: f(g(x)) -> g(g(g(x)))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: f#(g(x)) -> f#(x)

and R consists of:

r1: f(g(x)) -> g(g(f(x)))
r2: f(g(x)) -> g(g(g(x)))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        f#_A(x1) = max{1, x1 - 1}
        g_A(x1) = x1 + 4
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          f#_A(x1) = 0
          g_A(x1) = max{2, x1 - 3}
          f#!_A(x1) = 0
          g!_A(x1) = max{0, x1 - 3}
  
    partial status:
  
      pi(f#) = []
      pi(g) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.