First, consider 4 sequences $\left(x_{n}\right),\left(y_{n}\right),\left(z_{n}\right),\left(t_{n}\right)$ with $x_{1}=|a-b|$, $y_{1}=|b-c|$, $z_{1}=|c-d|$, $t_{1}=|d-a|$, $(a, b, c, d)$ is permutation of $(x, y, z, t)$. And
$$\left\{ \begin{array}{l} x_{n+1}=\left|a_{n}-b_{n}\right| \\ y_{n+1}=\left|b_{n}-c_{n}\right| \\ z_{n+1}=\left|c_{n}-d_{n}\right| \\ t_{n+1}=\left|d_{n}-a_{n}\right| \end{array} \right.$$
with $(a_{n}, b_{n}, c_{n}, d_{n})$ is a permutation of $(x_{n}, y_{n}, z_{n}, t_{n})$.
From this, it is easy to see that $x_{n}, y_{n}, z_{n}, t_{n} \geq 0$ for all $n$.
Let $w_{n}=\max \left\{x_{n}, y_{n}, z_{n}, t_{n}\right\}, n \geq 1$. We have
$$\begin{aligned} w_{n} & =\max \left\{x_{n+1}, y_{n+1}, z_{n+1}, t_{n+1}\right\} \\ & =\max \left\{\left|a_{n}-b_{n}\right|,\left|b_{n}-c_{n}\right|,\left|c_{n}-d_{n}\right|,\left|d_{n}-a_{n}\right|\right\} \\ & \leq \max \left\{a_{n}, b_{n}, c_{n}, d_{n}\right\}=\max \left\{x_{n}, y_{n}, z_{n}, t_{n}\right\}=w_{n} \end{aligned}$$
The equality occurs when there is at least one number among $a_{n}, b_{n}, c_{n}, d_{n}$ is equal to $0$.
Hence the sequence $(w_{n})$ is non-increasing. Suppose that $k$ is a positive integer such that $(x_{k}, y_{k}, z_{k}, t_{k})=(x, y, z, t)$, then $w_{1} \geq w_{2} \geq \cdots \geq w_{k}=w_{1}$ which implies that $w_{1}=w_{2}=\cdots=w_{k}$. So there is at least one number among $(x, y, z, t)$ equal to $0$. Based on the definition of the sequences, we can see that for each $i=1,2,3, \ldots, k$, the tuple $(x_{i}, y_{i}, z_{i}, t_{i})$ has some common properties as follows:
1. There is at least one number that is $0$.
2. There are two numbers equal.
3. In the permutation to obtain the next term, two equal numbers must be adjacent.
It is easy to check that 3 tuples as follows satisfy the given conditions:

i. If $(x, y, z, t)=(0,0,0,0)$.
ii. If $(x, y, z, t)=(a, a, 0,0)$ with $a>0$ and its permutation.
iii. If $(x, y, z, t)=(2a, a, a, 0)$ with $a>0$ and its permutation.

Indeed, when $(x, y, z, t)$ is one of the 3 above tuples, then the form of tuple $\left(x_{n}, y_{n}, z_{n}, t_{n}\right)$ will not change. Continue, we will prove that if $(x, y, z, t)$ is a tuple that satisfies the condition, then it must be one of the 3 above forms. (*)

Without loss of generality, we may assume that $x \geq y \geq z \geq t=0, x>0$. Since there are two numbers equal among 4 numbers, we consider some cases:

1. If $x=y$ then we have $(x, y, z, t)=(a, a, b, 0)$ with $a \geq b \geq 0, a>0$. Based on the 3rd property, $\left(x_{1}, y_{1}, z_{1}, t_{1}\right)$ can only be $(0, a-b, b, a)$. Apply the 2nd property, we have:
- If $a=a-b, b=0$, we have $(x, y, z, t)=(a, a, 0,0)$, which satisfies the given condition.
- If $a=b, a-b=0$, then we have $(x, y, z, t)=(a, a, a, 0)$ with $a>0$. So $\left(x_{1}, y_{1}, z_{1}, t_{1}\right)$ will be $(0,0, a, a)$. But from this, the form will not change, so we cannot obtain the original form anymore. This case doesn't satisfy.
- If $a-b=b$ then $a=2b$, we have $(x, y, z, t)=(2b, 2b, b, 0)$ and $\left(x_{1}, y_{1}, z_{1}, t_{1}\right)=(0, b, b, 2b)$. But the form of this tuple will not change and similarly, this case doesn't satisfy.

2. If $y=z$, then we have $(x, y, z, t)=(a, b, b, 0)$ with $a \geq b>0$. So $(x_{1}, y_{1}, z_{1}, t_{1})$ can be $(a-b, 0, b, a)$. Apply the 2nd property, we have:
- If $a-b=0, a=b$, we have $(x, y, z, t)=(a, a, a, 0)$, not satisfy.
- If $a-b=a, b=0$, we have the tuple $(x, y, z, t)=(a, 0,0,0)$ and $\left(x_{1}, y_{1}, z_{1}, t_{1}\right)=(a, 0,0, a)$. It is easy to see that this case doesn't satisfy.
- If $a-b=b$, we have $(x, y, z, t)=(2b, b, b, 0)$ which satisfies.

3. If $z=0$ then we have $(x, y, z, t)=(a, b, 0,0)$ with $a \geq b \geq 0$. So $(x_{1}, y_{1}, z_{1}, t_{1})$ can be $(a-b, 0, b, a)$. Apply the 2nd property, we have:
- If $a-b=0, a=b$, we have $(x, y, z, t)=(a, a, 0,0)$, satisfies.
- If $a-b=a, b=0$, we have $(x, y, z, t)=(a, 0,0,0)$, not satisfy.
- If $a-b=b$, we have the tuple $(x, y, z, t)=(2b, b, 0,0)$ and $\left(x_{1}, y_{1}, z_{1}, t_{1}\right)=(b, b, 0,2b)$. It is easy to see that this case doesn't satisfy.

Therefore, there are 3 tuples that satisfy the given condition as above.