Is Work done to Transfer Energy?
The example below is for a perfect lossless circuit with no resistance for the sake of illustration only. In reality this does not exist, but again the purpose is for simplicity.
If there is 1 Joule of energy stored in a capacitor and then that capacitor is connected to a coil to transfer and then store the 1 Joule of energy into the coil's magnetic field, is this work done and if so how much work was done?

In the diagram below, C1 is initially charged, and stores 1J of energy. Now we close S1.



simulate this circuit &ndash; Schematic created using CircuitLab
That puts a high voltage on D1 and L1. D1 is reverse biased so does not conduct. Current builds in L1 due to the voltage across its terminals, while VC1 the voltage on C1 falls. Eventually, VC1 gets to zero. There is now a large current flowing in L1, so VC1 tries to continue to fall, but now the (ideal) diode D1 starts to conduct, which prevents VC1 from going negative.
The voltage across L1 is zero, so the current doesn't change. As VC1 is zero, there is no energy stored in it. All the energy now resides in the current flowing through L1.
The capacitor has done 1J of work on the inductor

Taking it one step further, once the 1 Joule of energy is fully stored in the coil's magnetic field it then has to go somewhere.

No, it doesn't 'have to go somewhere', it can just stay there as long as the current keeps circulating.

... It then collapses and a diode is used to redirect the 1 Joule of energy to charge another capacitor.

No, it doesn't just 'collapse', but we can interrupt the current, by opening S2.
When we do that, current can now no longer flow through D1, and has to be sourced from D2 instead. D2 pulls current out of C2, making its voltage negative, and the voltage across C2 increases while the current in L1 falls. Eventually the current gets to zero. D2 stops conducting, it's now reverse biassed, the voltage on L1 falls to zero, so its current stays at zero. C2 is now storing the same energy that was stored in L1, and was stored in C1.
The inductor has done 1J of work on C2

What is the total amount of work that has been done in storing energy starting from the 1 Joule of energy stored in the first discharging capacitor, which is then stored in the coil's magnetic field, and then lastly ending up as stored energy in the second charged capacitor?

No work has been done opening and closing ideal switches, no work has been done heating zero resistances or making ideal diodes conduct, and no work has been done launching EM radiation from these ideal wires. In all cases, the component storing the energy has done exactly its stored energy amount of work on the component receiving the energy.