I am looking to see if it is possible to find a transformation $T_i(f(x))$ such that
$$T_1\left(e^x+e^{ix}+e^{-x}+e^{-ix}\right)=e^x-ie^{ix}-e^{-x}+ie^{-ix}$$ $$T_2\left(e^x+e^{ix}+e^{-x}+e^{-ix}\right)=e^x-e^{ix}+e^{-x}-e^{-ix}$$ $$T_3\left(e^x+e^{ix}+e^{-x}+e^{-ix}\right)=e^x+ie^{ix}-e^{-x}-ie^{-ix}$$
It might be easier to write also in terms of powers of $i$ here, instead. Thus
$$T_1\left(e^x+e^{ix}+e^{-x}+e^{-ix}\right)=i^4e^x+i^3e^{ix}+i^2e^{-x}+i^1e^{-ix}$$ $$T_2\left(e^x+e^{ix}+e^{-x}+e^{-ix}\right)=i^4e^x+i^2e^{ix}+e^{-x}+i^{2}e^{-ix}$$ $$T_3\left(e^x+i^1e^{ix}+e^{-x}+e^{-ix}\right)=e^x+ie^{ix}+i^2e^{-x}+i^3e^{-ix}$$
I have not studied transformations in a while and I don't know of what kind of mechanism could leave the exponential power in tact but simply alter the complex coefficent power. I have tried using substitution methods but nothing is getting me very far. For example, a simple substitution such as $x=-iy$ will simply permute the summands. But since the coefficient powers are also changing, you can't simply use a complex scalar here either.
I don't know if this is what you are looking for, but with $$ T_1(f) = f''', \quad T_2(f) = f'', \quad T_3(f) = f' $$ your given equations are satisfied.