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We know that $d/dx$ is the generator of translation in the sense that $$e^{ad/dx}f(x)=f(x+a)\tag{1}$$ which can be easily be proved from the Taylor series of $f(x+a)$.

Studying the very basics of conformal group/transformations suggest that the generator of scale transformation, at least in 3D, is given by $x_i\partial_i$ (summed over $i$). Can we assume that the generator of scale transformation is given by $x\frac{d}{dx}$ in 1D? If so, can we in a similar way prove that, the action of $e^{\lambda x \frac{d}{dx}}$ on a function $f(x)$ gives $f(e^{\lambda} x)$?

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Hint: Rewrite the dilation operator $$x\frac{d}{dx}~=~\frac{d}{d\ln |x|}$$ as a translation operator, and use OP's Taylor formula/translation eq. (1) to deduce that $$e^{\lambda x \frac{d}{dx}}f(x)~=~f(e^{\lambda}x).$$

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