Potential Energy of body and Earth system

The task would not be difficult in the latter case but would be slow. According to the newton's law of gravitation two bodies attract each other with a force of

$$overrightarrow{F}= GMm/R^{2}$$

Where M is mass of body and m is mass of the other body Now let mass of Earth to be M and mass of object m. So

$overrightarrow{F}_{earth}= GMm/R^{2}$........................(1)

$overrightarrow{F}=Moverrightarrow{a}$.............................(2)

so from 1 and 2 $$overrightarrow{a}=Gm/R^{2}$$ and for the object it will be $overrightarrow{a}=GM/R^{2}$ So as the mass of earth is very large hence the accleration of object will be very great compared to accleration of earth.

Also Work done by an external agent

$$ W= int_a^b ! overrightarrow{F}. , mathrm{d}overrightarrow{x}$$

Now as force and displacement both are equal so work done will also be equal

To move the Earth, we have to exceed the force of gravity to get it started. Because of its larger mass, its acceleration due to a given applied force will be less than that of the smaller object. So yes, it will be harder to move. The applied force can be broken into two parts: the force to oppose the gravitational force on the Earth and the extra force required to accelerate the Earth. The work done by this applied force in moving the Earth through a given distance will also consists of two parts: the work done against the gravitational force, which will increase the gravitational potential energy of the Earth-object system, and the work done by the rest of the applied force which increases the Earth's kinetic energy.

If we consider the case where the Earth is moved to a new fixed position without any kinetic energy there then all the work done goes into increasing the Earth-object potential energy. As @Pranav Aggarwal points out, the work done against gravity in moving either object a given distance is equal as the forces on each object are equal and the gravitational potential energy of the Earth-object system is increased by the same amount.

Your question's premise is that a larger force is required to move the earth distance d away from the smaller mass, compared to the force required to move the smaller mass distance d from the earth. There are three points you should keep in mind:

1. The force between the objects is proportional to the product of the masses, and by Newton's third law, are equal in magnitude. Thus the magnitude of the work path integral on each object will be the same for both objects. What does vary between the two objects is the gravitational field that each feels. Each object experiences a gravitational field proportional to the mass of the other object, so in the frame of the moving earth, the massive earth feels only a tiny field of the second object. But the magnitude of work is the same.
2. Inversely, the force each object experiences is the negative gradient or (slope in one dimension) of the potential energy of the system vs distance between the objects. For each object to feel the same magnitude force, the change in potential energy must be the same in both reference frames.
3. The two scenarios that you imagine, the earth fixed vs the earth moving, are different inertial frames of reference (assuming constant velocities in each frame). If you conducted an experiment to measure water pressure (which reports on how potential energy changes with height) at the end of a hose fixed to the ground and attached to a jug of water on an elevator, you would see the same increase in water pressure in the frame of reference of the ground and the frame of reference of the elevator.