This might be a very simple doubt, but I am not able to prove or disprove it rigorously.
Canetti's work on "Security and Composition of Multiparty Cryptographic Protocols: JoC 2000" allows us to prove results of the following form: Let $g, f_1, f_2, \cdots, f_m$ be $n$-party functions and $\pi$ can securely compute $g$ in $(f_1, \cdots , f_m)$ hybrid model. Let $\rho_i$ computes $f_i$ securely, then the composed protocol $\Pi$ by replacing the ideal call to $f_i$ by a subroutine call to $\rho_i$ securely evaluates $g$.
Does this result translate equally well if we consider Nash equilibrium or any game-theoretic equilibrium instead of secure computation? To be more precise, suppose instead of securely evaluating the function, we have a protocol that is Nash equilibrium in $(f_1, \cdots , f_m)$hybrid model, then can we construct a protocol that is Nash equilibrium in the real world?
What I mean by that a protocol is a Nash equilibrium in a setting X (X=ideal, hybrid, or real world) is the following. Suppose the strategy profile of player $i$ be $\Sigma_i$ for $i \in [n]$. Running the protocol is a (strict) Nash equilibrium with a strategy ($\sigma_1, \sigma_2, \cdots, \sigma_n) \in \Sigma_1 \times \cdots \times \Sigma_n$ in the setting X if any deviation of the players from the strategy prescribed in the protocol leads to a (strict) decrease in their utility.