1. Given: $f : \mathbb{R} \to \mathbb{R}$ is a continuous function. For any $c > 0$, the graph of $f$ can be moved to the graph of $cf$ using only a translation or a rotation.
2. Objective: Determine if this implies that $f(x) = ax + b$ for some real numbers $a$ and $b$.

Let's analyze the given condition in detail.

3. Translation: Consider the translation $(x, y) \mapsto (x - h, y)$. This means shifting the graph horizontally by $h$. If the graph of $f$ can be translated to the graph of $cf$, then there exists some $h$ such that:
$$f(x - h) = cf(x)$$

4. Rotation: Consider the rotation around the origin by an angle $\theta$. This means transforming the graph by rotating it. If the graph of $f$ can be rotated to the graph of $cf$, then there exists some $\theta$ such that:
$$f(x \cos \theta - y \sin \theta) = cf(x \sin \theta + y \cos \theta)$$

5. Example Function: The solution suggests that $f(x) = e^x$ also works. Let's verify this:
- For $f(x) = e^x$, consider the translation $(x, y) \mapsto (x - \ln c, y)$. This means shifting the graph horizontally by $\ln c$. Then:
$$f(x - \ln c) = e^{x - \ln c} = e^x \cdot e^{-\ln c} = e^x \cdot \frac{1}{c} = \frac{1}{c} e^x$$
- To match $cf(x)$, we need:
$$cf(x) = c e^x$$
- Therefore, the translation $(x, y) \mapsto (x - \ln c, y)$ maps the graph of $f$ onto the graph of $cf$.

6. Conclusion: Since $f(x) = e^x$ satisfies the given condition and is not of the form $ax + b$, the given condition does not necessarily imply that $f(x) = ax + b$.

The final answer is False.