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.
But perhaps you meant to ask whether one can always compute the isomorphism from an oracle for Tot, which is $\Pi_2$-complete and hence equivalent to an oracle for $0''$. In this case, the answer is also negative.
Let's start with the $0'$ case.
Theorem. There are two computable orders that are isomorphic, but the isomorphism is not computable from $0'$.
Proof. Let the first order be a natural copy of $\omega^2$, for which we can compute the function $(n,k)\mapsto\omega\cdot n+k$. Let the second order be as follows. Order the Turing machines $p$ in order type $\omega$. Create an interval in our new order associated with $p$. The interval will either be finite or infinite (but infinitely many of them will be infinite, and so the order overall will have type $\omega^2$). We simulate all programs on input $0$, input $1$, input $2$ and so on. Every time a program halts on the next input, we add another point to its interval block. Thus, the total programs will lead to infinite intervals, but the non-total programs will lead to finite intervals, since they will be waiting for their next input to halt.
There can be no $0'$-computable isomorphism from the first interval to the next, since from any such isomorphism, we could tell from $0'$ whether or not an interval was infinite or not, and thereby come to solve the $\Pi_2$-complete problem, which is not possible using only $0'$. $\Box$
I'll update more later with the $0''$ argument.