5. Notice that the sequence obtained after the exchange $t(A)$ also satisfies the inequality $0 \leqslant t\left(a_{i}\right) \leqslant i(i=0,1,2, \cdots, n)$. We will call all sequences that satisfy these inequalities as bounded index sequences. Below we prove $a_{i} \leqslant t\left(a_{i}\right)(i=0,1,2, \cdots, n)$.

If $a_{i}=0$, it is obviously true. Otherwise, let $x=a_{i}>0, y=t\left(a_{i}\right)$. The first $x$ terms $a_{0}, a_{1}, \cdots, a_{x-1}$ are all not greater than $x-1$, so they are all different from $x$, and they are before $x$ (as shown in Table 1), so $y \geqslant x$, i.e., $a_{i} \leqslant t\left(a_{i}\right)=(i=0,1,2, \cdots, n)$. This shows that the sequence will stabilize after a finite number of $t$ transformations, because in the bounded index sequence, the value of the $i$-th term does not exceed $i$. Next,

The value of the $i(0 \leqslant i \leqslant n)$-th term has the following properties: (i) it is an integer; (ii) it has an upper bound $i$; (iii) its value does not decrease during the $t$ transformation; (iv) once the value stabilizes under the $t$ transformation, it does not change.
This means that after no more than $n$ $t$ transformations, the sequence will stabilize.
Finally, we need to prove that starting from the initial bounded index sequence $A=\left\{a_{i}\right\}(i=0,1, \cdots, n)$, after no more than $n-1$ $t$ transformations, a stable sequence can be obtained. We prove this by mathematical induction on $n$.
When $n=1$, the two possible sequences are $\left(a_{0}, a_{1}\right)=(0,0)$ and $\left(a_{0}, a_{1}\right)=(0,1)$. Under the $t$ transformation, they are already stable.
Assume that any bounded index sequence $\left\{a_{i}\right\}, i=0,1,2, \cdots, n$ becomes a stable sequence under the $t$ transformation after no more than $n-1$ $t$ transformations. Consider $A=\left\{a_{i}\right\}, i=0,1,2, \cdots, n+1$, we prove that $A$ stabilizes after no more than $n$ $t$ transformations.

Assume it requires $n+1$ $t$ transformations. This is only possible when $a_{n+1}=0$, and each transformation increases the value of the $(n+1)$-th term by exactly 1. Under the $t$ transformation, the value of the $i$-th term is not affected by the terms after it. According to the induction hypothesis, the subsequence $A^{\prime}=\left\{a_{i}\right\}, i=0,1,2, \cdots, n$, stabilizes after no more than $n-1$ $t$ transformations. Because the $n$-th term stabilizes at $x(x \leqslant n)$ after no more than $\min \{x, n-1\}$ $t$ transformations, and the value of the $(n+1)$-th term, based on the current assumption, becomes $x$ exactly after $x$ $t$ transformations. Therefore, it can be concluded that the $(n+1)$-th term equals the value of the $n$-th term after no more than $n-1$ $t$ transformations.

However, in a sequence, if two consecutive terms have the same value, they remain equal and stable. Because the value of the $n$-th term stabilizes after no more than $n-1$ $t$ transformations, $A$ stabilizes after no more than $n-1$ $t$ transformations. This contradicts the assumption that it requires $n+1$ $t$ transformations. Therefore, at most $n$ $t$ transformations are needed to stabilize an $(n+1)$-term bounded index sequence.
To prove that if for some $i, a_{i}=t\left(a_{i}\right)(i=0,1,2, \cdots, n)$, then the $t$ transformation will not change the value of the $i$-th term. Consider the following two cases:
(1) If $a_{i}=t\left(a_{i}\right)=0$. This indicates that every term to the left of $a_{i}$ is 0. Then the first $i$ terms of $t(A)$ are also 0, and this repeats.
(2) If $a_{i}=t\left(a_{i}\right)=x>0$, then the first $x$ terms are all different from $x$. Because $t\left(a_{i}\right)=x$, so $a_{x}, a_{x+1}, \cdots, a_{i-1}$ are all equal to $x$. Therefore, $t\left(a_{j}\right)=x(j=x, x+1, \cdots, i-1)$. Another transformation $t$ will not change the value of the $i$-th term (as shown in Table 2).