Hardness of FNP vs Choice of Verifiers
No (unless $P=NP$).  Consider a language $L$ that is in $P$, say, $L$ is the language of prime numbers.

Suppose $V_1$ works as follows: $V_1(x,y_1)$ ignores $y_1$ and checks directly whether $x$ is prime, and if so, outputs 1; otherwise it outputs 0.

Suppose $V_2$ works as follows: $V_2(x,y_2)$ interprets $x$ as a 3CNF formula and checks whether $y_2$ is a satisfying assignment for it; then it checks whether $x$ is prime; and if both are true, it outputs 1, otherwise it outputs 0.

Both verifiers define the same language, namely, $L$.

However, given $x$, the problem of finding $y_1$ that's accepted by $V_1$ can be solved in polynomial time; but the problem of finding $y_2$ that's accepted by $V_2$ is NP-hard.  So, assuming $P \ne NP$, those two function problems have different hardness.