YES

Problem:
 f(f(x)) -> g(f(x))
 g(g(x)) -> f(x)

Proof:
 Arctic Interpretation Processor:
  dimension: 3
  interpretation:
             [1  1  1 ]  
   [g](x0) = [1  -& 0 ]x0
             [0  0  -&]  ,
   
             [0  1  -&]  
   [f](x0) = [0  1  0 ]x0
             [0  0  0 ]  
  orientation:
             [1 2 1]     [1 2 1]           
   f(f(x)) = [1 2 1]x >= [1 2 0]x = g(f(x))
             [0 1 0]     [0 1 0]           
   
             [2 2 2]     [0  1  -&]        
   g(g(x)) = [2 2 2]x >= [0  1  0 ]x = f(x)
             [1 1 1]     [0  0  0 ]        
  problem:
   f(f(x)) -> g(f(x))
  Arctic Interpretation Processor:
   dimension: 3
   interpretation:
              [0  0  1 ]  
    [g](x0) = [0  0  1 ]x0
              [-& -& -&]  ,
    
              [0 1 0]  
    [f](x0) = [2 1 1]x0
              [0 0 0]  
   orientation:
              [3 2 2]     [2  1  1 ]           
    f(f(x)) = [3 3 2]x >= [2  1  1 ]x = g(f(x))
              [2 1 1]     [-& -& -&]           
   problem:
    
   Qed