Suppose, $A$ is a well-ordered set with strict well-order $<_A$. Suppose $B$ is the set of all bijections from $A$ to $A$. Let’s define a relation $<_B$ on $B$ in the following way: $b <_B a$ iff $\exists c \in A$, such that $b(c) <_A a(c)$ and $\forall d <_A c$ $b(d) = a(d)$. Is $<_B$ always a strict well-order?
I know that it is always a strict total order, because it is both transitive and trichotomous. If $A$ is finite, then it is also a well-order. But what is about the general case?
On
Ivan Neretin gives a nice answer. I will show that your order is not well-ordered even if the ordertype of $A$ is $\omega$.
Consider the following sequences: $$a_n= \langle \underbrace{0,1,2,3,\cdots,2n-2,2n-1}_{\text{$2n$ terms}}, 2n+1, 2n, 2n+3,2n+2,\cdots\rangle$$
(The previous example is not bijective, so I changed the example.)
For example, $a_0=\langle1,0,3,2,5,4,\cdots\rangle$ and $a_1=\langle0,1,3,2,5,4,\cdots\rangle$. We can see that $a_0>a_1>a_2>\cdots$ under the lexicographic order.
OK, this is simply the lexicographic order on the permutations of A.
Say, A is the set corresponding to the ordinal $\omega+1$, that is, $A=\{1,2,3,\dots,\omega\}$. Now consider the following infinite series of permutations: $$\begin{aligned} a_1=&[\omega,2,3,4,\dots,1]\\ a_2=&[1,\omega,3,4,\dots,2]\\ a_3=&[1,2,\omega,4,\dots,3]\\ \vdots\\ \text{etc.} \end{aligned} $$ etc.
Clearly each $a_i$ is smaller than the previous, but where is the smallest element?