YES

We show the termination of R/S, where R is

  t(f(x),g(y),f(z)) -> t(z,g(x),g(y))
  t(g(x),g(y),f(z)) -> t(f(y),f(z),x)

and S is:

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

Since R almost dominates S and S is non-duplicating,
it is enough to show finiteness of (P, Q).
Here P consists of the dependency pairs

  t#(f(x),g(y),f(z)) -> t#(z,g(x),g(y))
  t#(g(x),g(y),f(z)) -> t#(f(y),f(z),x)

and Q consists of the rules:

  t(f(x),g(y),f(z)) -> t(z,g(x),g(y))
  t(g(x),g(y),f(z)) -> t(f(y),f(z),x)
  f(g(x)) -> g(f(x))
  g(f(x)) -> f(g(x))
  f(f(x)) -> g(g(x))
  g(g(x)) -> f(f(x))

The weakly monotone algebra (N^2, >_lex) with

  t#_A((x1,y1),(x2,y2),(x3,y3)) = (x1 + x2 + x3, y1 + y3)
  f_A((x1,y1)) = (x1 + 1, 1)
  g_A((x1,y1)) = (x1 + 1, 1)
  t_A((x1,y1),(x2,y2),(x3,y3)) = (x1 + x2 + x3, y2)

strictly orients the following dependency pairs:

  t#(f(x),g(y),f(z)) -> t#(z,g(x),g(y))
  t#(g(x),g(y),f(z)) -> t#(f(y),f(z),x)

We remove them from the set of dependency pairs.

No dependency pair remains.