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