To get some grip on the problem I considered the functions $f(x):=4x-x^2$ and $$g(x):=f\bigl(f\bigl(f(x)\bigr)\bigr)-x=63 x - 336 x^2 + 672 x^3 - 660 x^4 + 352 x^5 - 104 x^6 + 16 x^7 - x^8\ .$$
Plotting $g$ revealed 8 real zeros of g, two of them being $0$ and $3$ and the other six irrational.
Now obviously $x=y=z=0$ with sum $0$ and $x=y=z=3$ with sum $9$ solve the equations (but not the problem). Furthermore the data show that we can put
$$x=0.4679111137620438,\quad y=1.6527036446661105,\quad z=3.8793852415690395$$ with sum $6$ and
$$x=0.7530203962825327,\quad y=2.445041867912943,\quad z=3.801937735808046$$
with sum $7$.
Altogether this suggests that the possible sums $x+y+z$ are $0$, $6$, $7$, and $9$.
In order to attack the problem algebraically we introduce the elementary symmetric functions $s_1$, $s_2$, $s_3$ of $x$, $y$, $z$.
Adding the three equations gives
$s_1=4s_1-(s_1^2-2 s_2)$ or
$$s_2=(s_1^2-3s_1)/2\ ,\qquad(1)$$ and multiplying them gives
Since we may assume $s_3=xyz\ne0$ we can conclude
To get a third equation
we multiply the three equations of the form
and divide by $(y-z)(z-x)(x-y)\ne0$. Thus we get
which can be rewritten as
$$1=64-32 s_1+4s_1^2 + 4s_2-s_1 s_2+s_3\ .\qquad(3)$$
Eliminating $s_2$ and $s_3$ from $(3)$ by means of $(1)$ and $(2)$ finally gives the equation
which has the solutions found numerically before.