YES

Problem:
 f(g(X)) -> g(f(f(X)))
 f(h(X)) -> h(g(X))

Proof:
 String Reversal Processor:
  g(f(X)) -> f(f(g(X)))
  h(f(X)) -> g(h(X))
  Matrix Interpretation Processor: dim=3
   
   interpretation:
              [1 1 1]  
    [h](x0) = [0 0 0]x0
              [0 0 0]  ,
    
              [1 0 0]     [0]
    [f](x0) = [0 0 1]x0 + [0]
              [0 1 0]     [1],
    
              [1 0 0]  
    [g](x0) = [0 1 1]x0
              [0 1 1]  
   orientation:
              [1 0 0]    [0]    [1 0 0]    [0]             
    g(f(X)) = [0 1 1]X + [1] >= [0 1 1]X + [1] = f(f(g(X)))
              [0 1 1]    [1]    [0 1 1]    [1]             
    
              [1 1 1]    [1]    [1 1 1]           
    h(f(X)) = [0 0 0]X + [0] >= [0 0 0]X = g(h(X))
              [0 0 0]    [0]    [0 0 0]           
   problem:
    g(f(X)) -> f(f(g(X)))
   String Reversal Processor:
    f(g(X)) -> g(f(f(X)))
    Matrix Interpretation Processor: dim=3
     
     interpretation:
                [1 1 1]     [1]
      [f](x0) = [0 1 0]x0 + [0]
                [0 0 1]     [0],
      
                [1 0 0]     [0]
      [g](x0) = [0 1 1]x0 + [1]
                [0 1 1]     [1]
     orientation:
                [1 2 2]    [3]    [1 2 2]    [2]             
      f(g(X)) = [0 1 1]X + [1] >= [0 1 1]X + [1] = g(f(f(X)))
                [0 1 1]    [1]    [0 1 1]    [1]             
     problem:
      
     Qed