Solution:
We show that the solution set consists of $\{(t, 0, 0) ; t \in \mathbb{Z}\} \cup \{(-1, -1, 1)\}$. Let us put $a + b + c = d$, $ab + bc + ca = e$ and $abc = f$. The given condition $f(f(a, b, c)) = (a, b, c)$ implies that
$$d + e + f = a, \quad de + ef + fd = b, \quad def = c$$
Thus $abcdef = fc$ and hence either $cf = 0$ or $abde = 1$.

Case I: Suppose $cf = 0$. Then either $c = 0$ or $f = 0$. However $c = 0$ implies $f = 0$ and vice-versa. Thus we obtain $a + b = d$, $d + e = a$, $ab = e$ and $de = b$. The first two relations give $b = -e$. Thus $e = ab = -ae$ and $de = b = -e$. We get either $e = 0$ or $a = d = -1$.
If $e = 0$, then $b = 0$ and $a = d = t$, say. We get the triple $(a, b, c) = (t, 0, 0)$, where $t \in \mathbb{Z}$. If $e \neq 0$, then $a = d = -1$. But then $d + e + f = a$ implies that $-1 + e + 0 = -1$ forcing $e = 0$. Thus we get the solution family $(a, b, c) = (t, 0, 0)$, where $t \in \mathbb{Z}$.

Case II: Suppose $cf \neq 0$. In this case $abde = 1$. Hence either all are equal to $1$; or two equal to $1$ and the other two equal to $-1$; or all equal to $-1$.
Suppose $a = b = d = e = 1$. Then $a + b + c = d$ shows that $c = -1$. Similarly $f = -1$. Hence $e = ab + bc + ca = 1 - 1 - 1 = -1$ contradicting $e = 1$.
Suppose $a = b = 1$ and $d = e = -1$. Then $a + b + c = d$ gives $c = -3$ and $d + e + f = a$ gives $f = 3$. But then $f = abc = 1 \cdot 1 \cdot (-3) = -3$, a contradiction. Similarly $a = b = -1$ and $d = e = 1$ is not possible.
If $a = 1, b = -1, d = 1, e = -1$, then $a + b + c = d$ gives $c = 1$. Similarly $f = 1$. But then $f = abc = 1 \cdot 1 \cdot (-1) = -1$ a contradiction. If $a = 1, b = -1, d = -1, e = 1$, then $c = -1$ and $e = ab + bc + ca = -1 + 1 - 1 = -1$ and a contradiction to $e = 1$. The symmetry between $(a, b, c)$ and $(d, e, f)$ shows that $a = -1, b = 1, d = 1, e = -1$ is not possible. Finally if $a = -1, b = 1, d = -1$ and $e = 1$, then $c = -1$ and $f = -1$. But then $f = abc$ is not satisfied.
The only case left is that of $a, b, d, e$ being all equal to $-1$. Then $c = 1$ and $f = 1$. It is easy to check that $(-1, -1, 1)$ is indeed a solution.

Alternatively
$cf \neq 0$ implies that $|c| \geq 1$ and $|f| \geq 1$. Observe that
$$d^{2} - 2e = a^{2} + b^{2} + c^{2}, \quad a^{2} - 2b = d^{2} + e^{2} + f^{2}$$
Adding these two, we get $-2(b + e) = b^{2} + c^{2} + e^{2} + f^{2}$. This may be written in the form
$$(b + 1)^{2} + (e + 1)^{2} + c^{2} + f^{2} - 2 = 0$$
We conclude that $c^{2} + f^{2} \leq 2$. Using $|c| \geq 1$ and $|f| \geq 1$, we obtain $|c| = 1$ and $|f| = 1$, $b + 1 = 0$ and $e + 1 = 0$. Thus $b = e = -1$. Now $a + d = d + e + f + a + b + c$ and this gives $b + c + e + f = 0$. It follows that $c = f = 1$ and finally $a = d = -1$.