@TerryTao: Thanks! And thanks for not hesitating to post your partial results (which, as you can see, were a huge help).

That works for me.

Fair enough! I've always thought the name a bit unfortunate too.

It is possible to put a compact group topology on $\mathbb{R}$ (Halmos, "Comment on the real line", 1944). But any such topology is weird, in the sense that it cannot look very much like the usual topology on $\mathbb{R}$. For example, $[0,1)$ is not Haar measurable in any compact group topology on $\mathbb{R}$ (if it were, then you get a contradiction by the same argument that shows Vitali sets are non-measurable).

I have to admit that I won't really understand the link you give without putting a lot more effort in. That said, I'll share one more observation. If $A$ is such that there is an injective regressive function $\varphi: A \rightarrow \kappa$, then the answer to your original question is yes. (On the image of $\varphi$, define $f$ so that $f(\varphi(i)) \neq f_i(\varphi(i))$ for every $i \in A$. Off of the image of $\varphi$ define $f$ arbitrarily.) Since Fremlin mentions injective regressive functions in his argument, maybe this is relevant?

There are two trivial ways this can fail. If $0 \in A$ then $f_0$ and $f \upharpoonright 0$ are both the empty function (hence equal). A little less trivially, if $0 \notin A$ but $n \in A$ for every $0 < n < \omega$, then suppose $f_{n+1}$ is the function mapping $m$ to $0$ for $m < n$ and $n$ to $1$. If your condition is satisfied, this forces $f$ to map every $n < \omega$ to $0$. But if $f_\omega$ is constantly $0$, your condition fails at $\omega$. I'm writing this as a comment instead of an answer because I'm sure it's not really what you're wanting. Maybe put some restrictions on $A$?