Let $W$ be the Lambert W function, $Y$ be a real valued function and $x \in \mathbb{R} $.
Given $ Ye^Y = x \iff Y = W(x) $ is it true that $Y = kW(\frac{1}{k}x)$ for non-zero $k \in \mathbb{R} $ ?
I think this is true because: $$ Ye^Y = x \\ \iff \frac{Ye^Y}{k} = \frac{x}{k} \\ \iff \frac{1}{k} \cdot Ye^Y = \frac{x}{k} \\ \iff \frac{1}{k}Y = W\left(\frac{x}{k}\right) \\ \iff Y = k W\left(\frac{1}{k} \cdot x \right) \\ \square $$
No this proof does not hold, because this step is completely unjustified without assuming the theory therefore this is just circular logic: $$\iff \frac{1}{k} \cdot Ye^Y = \frac{x}{k} \\ \iff \frac{1}{k}Y = W\left(\frac{x}{k}\right) \\$$
The only way this step would be correct would be if the exponent ($Y$) was the same as the coefficient ($\frac1kY$)
A simple counterexample (let x = 5) $$W(5) = 1.326...$$ $$2 W(\frac125) = 0.590...$$
