Let $A$ be a set and suppose there are two relations on $A$, say $R$ and $S$, such that $(A,R)$ and $(A,S)$ are well-orders with the same order type, i.e. $(A,R)\cong(A,S)\cong(\alpha,\in)$ for some ordinal $\alpha$.
Is it true that that $R=S$?
I guess the answer is affermative. Anyway I can't provide a proof or a counterexample to refute this fact.
No, this not at all true, at least if $A$ contains at least two distinct elements.
Just notice that $\{a,b\}$ can be ordered as $a<b$ or $b<a$, and both are isomorphic to the unique ordinal $2$, but they are not equal.
More generally, given any permutation $\pi$ of $A$ and a well-ordering $R$, we can use $\pi$ to define $S$ to be $a\mathrel{S}b\iff\pi(a)\mathrel{R}\pi(b)$, which makes $S$ and $R$ isomorphic, but not equal, unless $\pi$ was the identity.
