How a uniform gravitational field affects the wavelength of a light beam?

Question

I was reading Einstein's 1911 paper named "On the Influence of Gravitation on the Propagation of Light" when is stated a formula for frequencies measured by observers at different fixed positions (heights) on Earth surface. One observer is at the origin of some coordinate system and measures a frequency $\nu_0$of a light beam. The second observer is at some position $x$ in the same chart and measures a frequency $\nu$ of the same light beam. Einstein obtais that

$$ \nu = \nu_0 \left( 1 + \frac{\phi}{c^2}\right) \tag1$$

where $\phi \le 0$ such that $\nu \le \nu_0$, the famous gravitational redshift.

However, at the end of section 3 of this paper, Einstein came to other result about speed of light measured by these observers:

$$ c= c_0 \left( 1 + \frac{\phi}{c^2} \right) \tag2$$

and so $c\le c_0$. Here $c$ is measured by the "$\nu$"  observer an its called "coordinate speed of light" and $c_0$ is measured by the "$\nu_0$" observer, and it is the usual vaccuum speed of light.

My question is: How the wavelength should transform from one observer to the other? I mean, if I take eq. (1), I could use

$$\frac{c}{\lambda} = \frac{c_0}{\lambda_0}\left( 1+ \frac{\phi}{c^2} \right) $$

Then by eq.(2) we obtain $\lambda =\lambda_0$,   which I think is non-sense because since the observers have two different frequencies of the same light beam, they certainly should disagree about its wavelength. Where is my mistake?