As the name implies, this question is somewhat similar in spirit to the previous question I asked with same title. This question is also about existence (or lack thereof) of certain possibilities concerning two different recursive well-orders with same order-type.

In my original question (https://mathoverflow.net/questions/278045/mapping-between-notations) I mentioned a result (in the comments below the original question) that is necessary to pose this particular question. I will use similar terminology as in the original question.

The idea behind this question goes as follows. Suppose we have two different well-orders of $\mathbb{N}$ with same order-type. Define the isomorphism function from the first well-order to second one as $P_{\,12}:\mathbb{N} \rightarrow \mathbb{N}$ (and $P_{\,21}:\mathbb{N} \rightarrow \mathbb{N}$ for the analogous function in the opposite direction).

Suppose that both $P_{\,12}$ is non-recursive but is $0'$computable. Then one of the following possibilities can hold:

 1. $P_{\,12}$ is recursively bounded and $P_{\,21}$ is $0'$computably bounded (but not recrusively bounded)  
 2. $P_{\,12}$ is $0'$computably bounded (but not recrusively bounded) and $P_{\,21}$ is recursively bounded
 3. $P_{\,12}$ is $0'$computably bounded (but not recrusively bounded) and $P_{\,21}$ is $0'$computably bounded (but not recrusively bounded)  

The possibility of $P_{\,12}$ and $P_{\,21}$ being both recursively bounded is ruled out because it would contradict $P_{\,12}$ being non-recursive.  

In a similar manner, suppose that $P_{\,12}$ is not $0'$ computably, but is $0''$ computable. Then only one of the following possibilities can hold:
 
 1. $P_{\,12}$ is recursively bounded and $P_{\,21}$ is $0''$computably bounded (but not $0'$ computably bounded)
 2. $P_{\,12}$ is $0''$computably bounded (but not $0'$ computably bounded) and $P_{\,21}$ is recursively bounded
 3. $P_{\,12}$ is $0'$ computably bounded (but not recursively bounded) and $P_{\,21}$ is $0''$computably bounded (but not $0'$ computably bounded) 
 4. $P_{\,12}$ is $0''$computably bounded (but not $0'$ computably bounded) and $P_{\,21}$ is $0'$ computably bounded (but not recursively bounded)
 5. $P_{\,12}$ is $0''$computably bounded (but not $0'$ computably bounded) and $P_{\,21}$ is $0''$computably bounded (but not $0'$ computably bounded)

For example, here the possibility of both $P_{\,12}$ and $P_{\,21}$ being $0'$ computably bounded is ruled out on the simple ground that a very simple generalisation of the result I have mentioned below shows that this would imply $P_{\,12}$ being $0'$ computable.

I guess there is a somewhat general pattern that emerges from this. So I guess the general question is that (for some natural number $n$) when $P_{\,12}$ is $0^{(n)}$ computable but not $0^{(n-1)}$ computable, then which of the possibilities (amongst the $2n+1$) can or can't occur? 

**P.S.** Adding link to the argument for the mentioned result (and justifying the exclusion of possibilities ignored in the question):
(i) [Main Argument][1] (ii) [Program Implementation][2]

The functions that are used in program implementation:
(iii) [Interface/List of Elementary Functions][3]  (iv) [Implementation][4] 

It is link (i) that is really relevant here (and to some extent (ii) possibly). I am posting links (iii) and (iv) just for the case if someone might be interested in following (ii) .... so the interface and/or implementation of specific functions can be looked up in a transparent way if necessary. 

I wrote (i) and (ii) about nine months ago and (iii) and (iv) about little over one and a half years ago. So the writing might not be without mistakes (though at that time I did re-check (i) a couple of times .... especially for more substantial argument-breaking mistakes). Also there are occasions of slightly odd terminology, which I didn't define. Clarifying it here: (a) $PS$ means $\omega_{CK}$ (b) I have written $N_1/p$ and $N_2/p$ for "recursive numberings". What that simply means is two different recursive well-orders for $\mathbb{N}$ with order-type $p+1$. Only condition is that the number $0$ must be assigned to $p$ (that is $address1(p)=0$ and $address2(p)=0$).  

  [1]: https://ln.sync.com/dl/b21d424f0/q6sm428m-xg4r3m7r-5y5zh62j-myin58ww 
  [2]: https://ln.sync.com/dl/81f221710/y45ng85s-dqbvpre8-swfkg49r-u5pzzh54
  [3]: https://ln.sync.com/dl/b8e406970/yqipiy9g-9vctdhj2-6qpjeyk6-2c6h8ej9
  [4]: https://ln.sync.com/dl/9eb7a8460/cskux3p7-eia74rgs-wyzu6gnw-33ebxdc8

**Edit:** Updated document(i) by adding explanation of the (naive) program/algorithm, and also correcting a few mistakes. Main question remains unchanged.