I understand that when converting a CFG to a CNF that we need to add a new start symbol if the start symbol occurs on the right-hand side. For instance, this grammar is converted like such:
$$S\rightarrow ASA | a $$ $$A\rightarrow ab | ba | aa | bb | \epsilon $$
becomes
$$S_{0}\rightarrow S $$ $$ S\rightarrow ASA | a$$ $$A\rightarrow ab | ba | aa | bb | \epsilon $$
What if, however, there is a CFG that does not have the start symbol on the RHS of the equation? For instance, consider this equation:
$$S\rightarrow a | aA | B $$ $$A\rightarrow aBB | \epsilon $$ $$B\rightarrow Aa | b $$
As the first step of the problem, the source I found this example in insists that a new start symbol needs to be created immediately. I don't see why this is necessary given that the start symbol is not on the RHS.
Is it always necessary to create a new start symbol or is this just an extra step in this particular problem?
Furthermore, consider a CFG such as
$$S\rightarrow A$$ $$A\rightarrow aB | A | \epsilon$$ $$B\rightarrow c | ab$$
Does a new start symbol need to be added if there is no start symbol on the RHS and the start symbol simply points to some other rule?
Original source of problem in question: https://www.javatpoint.com/automata-chomskys-normal-form
No, you don't need to add a new start symbol unless the old start symbol occurs on the right hand side.
Chomsky normal form requires that productions all have the form $A\to BC$, $A\to a$, or $S\to \varepsilon$ (if the language contains the empty string $\varepsilon$), and requires that $S$ never occur on the right hand side.
Roughly speaking, there's a recipe for transforming your grammar into CNF. That recipe introduces some fresh new symbols on the right hand side, and in one case (eliminating $\varepsilon$ productions) creates new arrangements of the symbols that are already on the right hand side.
But these steps never put the $S$ symbol on the right hand side if it wasn't already there. So if your grammar already satisfies the "no start symbol on the right hand side" requirement, the remaining CNF conversion steps will keep it that way. You don't need a new start symbol to satisfy all the requirements.
(The worked example on that page is mistaken: $S$ doesn't occur anywhere on the right hand side, so accordingly you don't need a new start symbol.)