I understand your question as asking: given two isomorphic
computable well-orders, must there be a computable isomorphism?

The answer is no.

For a counterexample, let the first order be the natural numbers
with the usual ordering $\langle\mathbb{N},<\rangle$, which has
order type $\omega$. Let the second order be constructed in the
following computable manner. Put the Turing machines in order in
type $\omega$ and for each machine $p$, create two points $a_p$ and
$b_p$ and specify $a_p<b_p$, with each pair being ordered by the machine order. Now, begin simulating all programs,
and whenever a new program halts, add a new point $c_p$ with
$a_p<c_p<b_p$. This specifies a computable order with order type
$\omega$, and we can arrange that these points exhaust $\mathbb{N}$. But there can be no computable isomorphism between this
order and the first order, because if there were, we could
computably determine whether or not $b_p$ was a successor of $a_p$
or not, and thereby computable solve the halting problem, which is
impossible.

Conclusion: there are computable copies of $\omega$, with no
computable order-isomorphism between them.