When does an extendible 1:1 p.c. function have a 1:1 computable extension?

A partial computable function $\varphi_e$, defined on a c.e. set $W_e$, is called extendible if there exists some computable function $f$ which extends $\varphi_e$, i.e. $\varphi_e(W_e) = f(W_e)$.
My question is what are the conditions for a p.c. function $\varphi_e$ to have an injective extension.

So far I have found some necessary conditions:

1. Obviously, $\varphi_e$ itself should be injective.

2. $|\overline{W_e}| \leq |\overline{\varphi_e(W_e)}|$. Because $f(\overline{W_e})$ should have no intersection with $f(W_e) = \varphi_e(W_e)$, so $f(\overline{W_e}) \subseteq \overline{\varphi_e(W_e)}$.
This rules out $\varphi_e(x) = x/2$, defined on even numbers. And also any other surjective partial function.

I couldn't manage to find a sufficient condition yet.

Here is an example of an extendible, non-injectively-extendible p.c. function $\psi$ which nonetheless satisfies your conditions (and I think indicates in general that any sufficient condition will be quite complicated):
Let $\langle\cdot,\cdot\rangle$ be your favorite computable bijection $\mathbb{N}^2\rightarrow\mathbb{N}$. We define $\psi$ as follows (with the role of "$2^k$" being merely to control the range of $\psi$, and consequently show that just demanding that the range be "small" won't yield a sufficient condition):
• Suppose $k$ is even. Let $a_k, b_k$ be the unique natural numbers such that $k=2\langle a_k,b_k\rangle$. We define $\psi(k)$ to be $2^k$ if $\varphi_{a_k}(a_k)$ halts strictly before $\varphi_{b_k}(b_k)$ does, and we let $\psi(k)$ be undefined otherwise.
• Suppose $k$ is odd. Let $a_k, b_k$ be the unique natural numbers such that $k=2\langle a_k,b_k\rangle$+1. We define $\psi(k)$ to be $2^{k-1}$ if $\varphi_{b_k}(b_k)$ halts strictly before $\varphi_{a_k}(a_k)$ does, and we let $\psi(k)$ be undefined otherwise.
$\psi$ is clearly 1-1, p.c., of co-infinite range (indeed, the range is "computably coinfinite" in the sense that the complement of the range contains an infinite computable set; in jargon, the range of $\psi$ is non-simple), and extendible: it is extended by the function $x\mapsto 2^{2floor({x\over 2})}$. However, it's easy to show that it's not extendible to a 1-1 total computable function, essentially since such an extension would have to always guess correctly whether one program halts before another does.