Consider objects $A, B$ in some category $C$. Let $S\in C$ to be the sum of $A$ and $B$ with inclusions as $i_A:A\rightarrow S$ and $i_B:B\rightarrow S$.
If $A\cong B$ with isomorphisms $f:A\rightarrow B$ and $f^{-1}:B\rightarrow A$, then is there a map on $S$, $\alpha$, such that $$\alpha \circ i_A\circ f^{-1}=i_B \qquad \text{and}\qquad \alpha\circ i_B\circ f=i_A?$$
(In case of category set, it appears to be true, though I certainly doubt that it would be true in such a general context. Perhaps, someone more mature than me would be able to provide a counter example.)
Other related question is that if the above were true then is it true that $\alpha$ is an involution on $S$.
This is true in any category in which the coproduct/sum $A \amalg B$ exists.
Take your situation, where $S$ is the coproduct of $A$ and $B$ with structure maps $i_A: A \to S$ and $i_B: B \to S$. Apply the universal mapping property of the coproduct to the maps $i_B \circ f:A \to S$ and $i_A \circ f^{-1}: B \to S$. By universality, there is a unique map $\alpha: S \to S$ such that $\alpha \circ i_A = i_B \circ f$ and $\alpha \circ i_B = i_A \circ f^{-1}$. Hit the first relation with $f^{-1}$ on the right and the second with $f$ on the right to get your desired equalities.
Moreover, $\alpha$ will be an automorphism satisfying $\alpha \circ \alpha =1_S$, so it will typically not be an idempotent. It is easy to see that the identity map $1_S: S \to S$ is the unique map $x$ satisfying $x \circ i_A = i_A$ and $x \circ i_B = i_B$. But it is an easy exercise using the relations in the previous paragraph to show that these relations hold with $x = \alpha \circ \alpha$. By uniqueness, $\alpha \circ \alpha = 1_S$.
(A map $\alpha$ with $\alpha^2 =1$ is called an involution. Is this what you meant instead of "idempotent"?)