On the computable function of a problem that halts Let's say program $P$ with given input $i$ is found to halt (or doesn’t halt) by a Turing machine. Is it true that the same program $P$ with input $F(i)$ also halts (or not, respectively), where $F$ is a computable function?

In general, is it true that the behavior of program $G(P)$ with the input $F(i)$ identical to the behavior of P with input i? (Both $G$ and $F$ are computable.)
No, it is not.

In general, to help you think about this kind of problem, I recommend you work through some examples.  Keep them as simple as possible.

Here is an example that illustrates why the answer is no.  Suppose $P$ is the following program: "if the input is 1, then halt immediately; otherwise enter an infinite loop".  Suppose $F(1)=2$.  Then $P$ halts on the input $i=1$, but it does not halt on the input $F(i)=F(1)=2$.  So, the behavior of the program $P$ on one input tells you nothing about its behavior on a different input.