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, >) with

  t#_A(x1,x2,x3) = x1 + x2 + x3
  f_A(x1) = x1 + 1
  g_A(x1) = x1 + 1
  t_A(x1,x2,x3) = 0

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.