Is it true that $f$ must be constant? Let $f: \Bbb R \to \Bbb R$ be a function which satisfies $f(x) = f(2x-1)$ for all $x \in \Bbb R$ and $f$ is continuous at $x=1$. Is it true that $f$ must be constant?

From your assumption we have $f(x)=f(\frac{x}{2}+\frac{1}{2})$ for all $x$. For $x\in \mathbb R$ and substituting successively $x$ by $\frac{x}{2}+\frac{1}{2}$, we get
$$f(x)=f\Big(\frac{x}{2}+\frac{1}{2}\Big)=f\Big(\frac{1}{2}\Big(\frac{x}{2}+\frac{1}{2}\Big)+\frac{1}{2}\Big)=f\Big(\frac{x}{2^2}+\frac{1}{2}+\frac{1}{2^2}\Big)=\cdots =f(\frac{x}{2^n}+\frac{1}{2}+\cdots \frac{1}{2^n}\Big).$$
Note that $\frac{x}{2^n}+\frac{1}{2}+\cdots +\frac{1}{2^n} \to 1$ as $n\to \infty$ with fixed $x$. Then by the continuity of $f$ at $1$ we get $f(x)=f(1)$ for all $x$.
