How does the metric change under scale transformations?

I was reading Zee's book in Group Theory, chapter VIII.2 (The Conformal Algebra) and I'm stuck in something probably easy:

In p.516, he states that under a scale transformation, $x'^{\mu} = \lambda x^{\mu}$, the metric changes as:

$g'_{\rho\sigma}(x') = g'_{\rho\sigma}(\lambda x) = \lambda^2g_{\mu\nu}(x)$

However, this result seems wrong to me. First, the indices are wrong and using the general formula for transforming the metric I get:

$g'_{\rho\sigma}(x') = g_{\mu\nu}(x)\frac{\partial x^{\mu}}{\partial x'^{\rho}}\frac{\partial x^{\nu}}{\partial x'^{\sigma}} = \frac{g_{\mu\nu}(x)}{\lambda^2}\frac{\partial x^{\mu}}{\partial x^{\rho}}\frac{\partial x^{\nu}}{\partial x^{\sigma}} = \frac{g_{\mu\nu}(x)}{\lambda^2} \delta^{\mu}_{\rho}\delta^{\nu}_{\sigma} = \frac{g_{\rho\sigma}(x)}{\lambda^2}$

Is something wrong in my logic?

I think he means something more mundane i.e $$g_{\mu \nu} dx^{\mu}dx^{\nu} \rightarrow g_{\mu \nu} d (\lambda x^{\mu})d( \lambda x^{\nu}) = g_{\mu \nu} \lambda^2 dx^{\mu}dx^{\nu}$$. Then we can define new components $$g_{\mu \nu}' =g_{\mu \nu} \lambda^2$$. I think the confusion comes from the fact that we are not actually talking about a coordinate transformation (at least not in the usual sense). Consider the following: I have basis set $$\{e_i\}$$ and $$\{e_j\}$$. We can write a vector in each as follows $$v=v^ie_i$$ or $$v=v^je_j$$. There is a linear transformation $$A_{ij}e_j = e_i$$. This means $$v = v^iA_{ij}e_j$$ and therefore switching from the $$\{e_i\}$$ basis to $$\{e_j\}$$ requires multiplication of $$v$$ by $$A^T$$ or orthogonal transformations specifically $$A^{-1}$$.Something like this is what you have in mind with the equations you wrote.
I think what Zee has in mind is the following. In a specific coordinate system $$\{e_a\}$$, we can write the components of the covariant tensor as $$g_{ab}=e_a \cdot e_b$$. But in this coordinate system perform a scaling transformation. What do the matrix elements of the matrix tensor look like? Well, $$e_a \rightarrow \lambda e_a$$ so $$g_{ab}' = \lambda e_a \cdot \lambda e_b = \lambda^2 e_a \cdot e_b = \lambda^2 g_{ab}$$