Is there any characterization for the real-valued function such that $y - x = y' - x' \Rightarrow f(y) - f(x) = f(y') - f(x')$

I'm looking for a real-valued function, $f: \mathbb{R} \rightarrow \mathbb{R}$, which is monotone, i.e., $x \leq y \Rightarrow f(x) \leq f(y)$, and with a specific characterization which can be called constant equispaced intervals, it is expressed as: $$y - x = y' - x' \Longrightarrow f(y) - f(x) = f(x') - f(y')$$

Edit: It should be said that $f(x)$ is not differentiable necessarily.

I hope to find (if it is possible) an analytical expression of $f(x)$, may be with two or three parameters. But, if it isn't possible, at least a characterization of all these functions.

To solve this question, I tried two attempts:

1. First, finding a functional equation and trying to solve it: As $y - x = y' - x'$, then $f(y - x) = f(y' - x')$. Therefore, we have the following equation: $$\frac{f(y - x)}{f(y' - x')} = \frac{f(y) - f(x)}{f(y') - f(x')}$$

But, I don't know if this is helpful for something, I got stacked here.

2. Second, trying to solve a recursion with natural numbers: Consider the bivariate function $F(x, y) = f(y) - f(x)$. And let us fix the variable $y$ with a natural number $N$, which we can suppose very high. Then, we have: $F(2, N) - F(1, N) = F(3, N) - F(2, N) = \cdots = F(x+1,N) - F(x, N)$, and we denote this common difference with $b(N)$.

Thus, $F(2, N) = F(1, N) + b(N)$

$F(3, N) = F(2, N) + b(N) = F(1, N) + 2 \cdot b(N)$

$\vdots$

$F(x, N) = F(x-1, N) + b(N) = F(1, N) + (x-1) \cdot b(N)$

It seems that we have extracted the variable x from the functional dependence. But, I don't know if it is useful for something.

Any ideas will be grateful.

Let $g(x)=f(x)-f(0)$. Then $g$ satisfies the same property, but $g(0)=0$.

For any $n\in\mathbb N$, $q\in\mathbb R$, $g(nq)-g((n-1)q)=g(q)-g(0)=g(q)$. So \begin{align}\tag1 g(nq)&=g((n-1)q)+g(q)=g((n-2)q)+2g(q)\\[0.3cm]&=\cdots=g(q)+(n-1)g(q)=ng(q). \end{align} Taking $q=1/n$, we obtain $g(1)=ng(1/n)$, so $$\tag2g(1/n)=g(1)/n.$$ Combining $(1)$ and $(2)$, $$\tag3 g(\tfrac mn)=m\,g(\tfrac1n)=\tfrac mn\,g(1). $$ Because $f$ is monotone so is $g$, so it is continuous almost everywhere and any possible discontinuity is a jump one. But $(3)$ precludes jump discontinuities, so $g$ is continuous, and $$\tag4 g(t)=t\,g(1),\qquad t\in\mathbb R. $$ Thus $$\tag5f(t)=g(t)+f(0)=t(f(1)-f(0))+f(0). $$

HINT:

$$f(x+y) - f(x) = f(y) - f(0)$$ or $$f(x+y) - f(0) = (f(x)-f(0)) + (f(y) - f(0))$$ that is, the function $x\mapsto f(x) - f(0)$ is additive.