We say that a function $f: \mathbb{R} \rightarrow \mathbb{R}$ preserves the binary relation $\sim \subseteq \mathbb{R}^2$ if $x \sim y$ implies $f(x) \sim f(y)$ for all $x,y\in\mathbb{R}$.
We say that $\sim$ ensures continuity if every function $f: \mathbb{R} \rightarrow \mathbb{R}$ that preserves $\sim$ is continuous.
Part 1. Is there any binary relation that ensures continuity?
A well-ordering would work for this, I think (edit: Silly me, I realized this is not actually true. But a totally rigid relation exists on any set given Choice anyway.) But I'd like to see something better (in Part 2), and if possible something that doesn't require Choice for Part 1.
Part 2. Is there any binary relation that ensures continuity, and is preserved by at least two different functions?
This answer was simplified by a lot from the original one.
Consider a relation defined as follows: $x \sim y$ if $x > y$, or $y = x + 2^{m}$ for some $m\in\mathbb{Z}$.
The idea is that the first part of the definition will force the function to be strictly increasing, the second part constrains what the values of $f(x+2^m)$ might be in terms of $f(x)$. This then gives us a single sequence along which $f(x+2^m)$ converges to $f(x)$ and monotonicity then forces $f$ to be sequentially-continuous.
First, let's demonstrate that any function preserving this relation is strictly increasing. Assume $x > y$, if $f(x) \leq f(y)$ then we must have $f(y) = f(x) + 2^{m}$.
Pick $z$ such that $x>z>y$, then we have $f(x) \sim f(z) \sim f(y)$. There are uncountably many such $z$ so there must be uncountably many $z$ satisfying $f(x) > f(z)$, since otherwise we would have $f(z) = f(x) + 2^{m}$ for some $m(z)$. This implies $m(z_1) = m(z_2)$ for uncountably many pairs $z_1, z_2$. Since $f(z_1) \sim f(z_2)$ as well as $f(z_1) = f(z_2)$, we have a contradiction, as $a\sim a$ never holds for our relation.
We apply this same argument to all $z$ satisfying $f(x) > f(z)$, to establish $f(z) > f(y)$ for some $z$. Therefore, we have successfully shown $f(x) > f(y)$.
Next, let's demonstrate that any function preserving this relation is continuous. Since $x \sim x + 2^{m}$, we have $f(x) \sim f\left(x+2^{m}\right)$. As $f$ is strictly increasing, this can only occur if $f\left(x+2^{m}\right) = f(x) + 2^{k(m)}$. Therefore, $\lim_{m\to-\infty} f\left(x+2^{m}\right) = f(x)$, as $k(m)$ is strictly increasing. Consequently, $f$ being strictly increasing gives us right continuity.
Similarly, since $x - 2^{m}\sim x$, we have $f\left(x - 2^{m}\right) \sim f(x)$. Again, considering $f$ is strictly increasing, this can only occur if $f\left(x - 2^{m}\right) = f(x) - 2^{k(m)}$. Thus, $\lim_{m\to-\infty} f\left(x - 2^{m}\right) = f(x)$, since again $k(m)$ is strictly increasing. This leads to left continuity.
So we have shown that $f$ is continuous. Furthermore, $f(x) = 2^n x$ for any $n$ preserves this relation.