YES

Problem:
 h(x) -> g(x)
 g(x) -> h(x)
 f(x,y) -> f(y,x)
 f(g(x),h(y)) -> f(x,y)
 f(h(x),h(y)) -> f(y,x)

Proof:
 AT confluence processor
  Complete TRS T' of input TRS:
  f(g(x),h(y)) -> f(x,y)
  f(h(x),h(y)) -> f(y,x)
  f(g(x),g(y)) -> f(y,x)
  f(h(y),g(x)) -> f(y,x)
  h(x) -> g(x)
  g(x) -> h(x)
  f(x,y) -> f(y,x)
  
   T' = (P union S) with
  
   TRS P:h(x) -> g(x)
         g(x) -> h(x)
         f(x,y) -> f(y,x)
  
   TRS S:f(g(x),h(y)) -> f(x,y)
         f(h(x),h(y)) -> f(y,x)
         f(g(x),g(y)) -> f(y,x)
         f(h(y),g(x)) -> f(y,x)
  
  S is linear and P is reversible.
  
   CP(S,S) = 
  
   CP(S,P union P^-1) = 
  f(x482,x483) = f(h(x483),g(x482)), f(x484,x485) = f(h(x485),g(x484)), 
  f(x487,x486) = f(h(x487),h(x486)), f(x489,x488) = f(h(x489),h(x488)), 
  f(x491,x490) = f(g(x491),g(x490)), f(x493,x492) = f(g(x493),g(x492)), 
  f(x494,x495) = f(g(x495),h(x494)), f(x496,x497) = f(g(x497),h(x496))
  
   CP(P union P^-1,S) = 
  f(g(x),g(y)) = f(x,y), f(g(x),h(y)) = f(y,x), f(h(x),g(y)) = f(y,x), 
  f(g(y),g(x)) = f(y,x), f(h(x),h(y)) = f(x,y), f(h(y),h(x)) = f(y,x), 
  f(h(y),g(x)) = f(x,y)
  
  
  We have to check termination of S:
  
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [g](x0) = 6x0 + 4,
    
    [h](x0) = 7x0,
    
    [f](x0, x1) = x0 + 4x1
   orientation:
    f(g(x),h(y)) = 6x + 28y + 4 >= x + 4y = f(x,y)
    
    f(h(x),h(y)) = 7x + 28y >= 4x + y = f(y,x)
    
    f(g(x),g(y)) = 6x + 24y + 20 >= 4x + y = f(y,x)
    
    f(h(y),g(x)) = 24x + 7y + 16 >= 4x + y = f(y,x)
   problem:
    f(h(x),h(y)) -> f(y,x)
   Matrix Interpretation Processor: dim=1
    
    interpretation:
     [h](x0) = 3x0 + 1,
     
     [f](x0, x1) = 2x0 + x1
    orientation:
     f(h(x),h(y)) = 6x + 3y + 3 >= x + 2y = f(y,x)
    problem:
     
    Qed