Why does de Broglie wavelength lambda = c / frequency?
$v$ is not the wave speed, it is the speed of the object with the given mass and de Broglie frequency. Also, there should be rather $p$ instead of $mv$, unless $v \ll c$.
As $$ E = \sqrt{(mc^2)^2 + (pc)^2}$$
then
$$p = \sqrt{\left({\frac {E}{c}}\right)^2 - {(mc)}^2}$$
For a massless object like a photon:
$$p = \frac Ec$$.
For an object with mass, there is also $$p  = \frac {mv}{\sqrt{1-(v/c)^2}}= \gamma mv$$
See Matter wave, paragraphs De Broglie relations and Special relativity, where is clear c is the phase speed of light in vacuum:

$$\begin{align}&amp;\lambda =\,\, \frac {h}{\gamma m_0v}\, =\, \frac {h}{m_0v}\,\,\,\, \sqrt{1 - \frac{v^2}{c^2}}\\
&amp; f = \frac{c}{\lambda} = \frac{\gamma\,m_0vc}{h} = \frac {m_0c^2}{h} \bigg/ \sqrt{\frac{c^2}{v^2}-1}
\end{align}$$