When two identical bodies are brought into thermal contact, and both are surrounded by an isolation, does the fact that they have a finite temperature difference, which generates entropy, change their final temperatures. Are their final temperatures their arithmetic average
Let's consider a bathtub full of hot water @ $T_1 = 35°C$ and a "small" stone @ temperature $T_2 = 5°C$. From this setup is should be clear that if we bring the two "bodies" together, the final temperature will not be their arithmetic average, $\bar T = (T_1 + T_2)/2 = 20°C$, but that it is much closer to $35°C$.
Unfortunately, your actual question
... does the fact that [... the two bodies] have a finite temperature difference [...] change their final temperatures?
I do not understand the question. The final temperature depends on
- the initial temperature of each body,
- the specific heat capacity of each body,
- the mass of each body.
The idea is that heat (=energy) flows from the hotter body to the colder body. The energy drain reduces the temperature of the hotter body according to its heat capacity. Analogously, the energy gain increases the temperature of the colder body. The heat flow effectively stops if the two bodies are in thermal equilibrium (=same temperature). Although we still expect to obtain "small" fluctuations of the temperatures these are tiny. The argument uses (1) the equal a priori probability law and (2) the fact that the number of accessible micro-states strongly depends on the temperature of each body.