De-Broglie wavelength from two different reference frames If we were to measure the De-Broglie wavelength of a particle of mass $m$ and velocity $v$, in a given reference frame, it would be given as $$\lambda = h/p$$
(where $h$ is Planck's constant & $p$ is momentum)
Now if we were to measure the De-Broglie wavelength of the same particle from a reference frame moving with velocity $v$ the particle would seem stationary, so $p=0$ and its De-Broglie wavelength would be $$\lambda = h/0$$
So how exactly does a particle possess two different wavelengths when observed from two different reference frames?
 A: The de Broglie frequency and wavelength of the "superluminar" plane phase wave accompaning a "subluminar" particle moving with speed $\:\upsilon\:$ are(1)
\begin{equation}
\nu \boldsymbol{=} \dfrac{E }{h} \boldsymbol{=} \dfrac{\sqrt{\left(m_{\mathrm{o}}c^{2}\right)^{2} \boldsymbol{+}\left(pc\right)^{2}}}{h} \boldsymbol{=}\dfrac{\gamma m_{\mathrm{o}}c^{2}}{h}
\tag{01}\label{01}  
\end{equation}
\begin{equation}
\lambda  \boldsymbol{=} \dfrac{h }{p} \boldsymbol{=}\dfrac{h}{\gamma m_{\mathrm{o}}v}
\tag{02}\label{02}  
\end{equation}
where $\:\gamma=\left(1\boldsymbol{-}\dfrac{\upsilon^2}{c^2}\right)^{\boldsymbol{-}1/2}\:$ and $\:m_0\:$ the rest mass.
The plane phase wave "superluminar" speed is
\begin{equation}
\mathrm w=\lambda\,\nu \boldsymbol{=} \dfrac{c^{2}}{\upsilon}
\tag{03}\label{03}  
\end{equation}
In the rest frame  of the particle
\begin{equation}
\nu_{0} \boldsymbol{=} \dfrac{E_{0}}{h} \boldsymbol{=} \dfrac{m_{\mathrm{o}}c^{2}}{h}
\tag{04}\label{04}  
\end{equation}
\begin{equation}
\lambda_{0}  \boldsymbol{=} \infty
\tag{05}\label{05}  
\end{equation}
\begin{equation}
\mathrm w_{0}  \boldsymbol{=} \infty
\tag{06}\label{06}  
\end{equation}
that is a uniform in space periodic in time vibration with frequency $\:\nu_{0}$.

EDIT :
responds to the OP's comment

I still don't know a lot of stuff in physics, the only thing I knew in your answer was the square root of the Lorentz gamma. Do you think you maybe dumb it down a little? – Kosh Rai

De Broglie hypothesis : 

$^{\prime\prime}$The fundamental idea of [my 1924 thesis] was the following : The fact that, following Einstein's introduction of photons in light waves, one knew  that  light  contains  particles  which  are  concentrations of energy incorporated into the wave, suggests that all particles, like the electron, must be transported by a wave into which it is incorporated... My essential idea was to extend to all particles the coexistence of waves and particles discovered by Einstein in 1905 in the case of light and photons.$^{\prime\prime}$ $^{\prime\prime}$With every particle of matter with mass $m$ and velocity $\upsilon$ a real wave must be 'associated'$^{\prime\prime}$, related to the momentum by the equation:
  \begin{equation}
\lambda  \boldsymbol{=} \dfrac{h}{p}\boldsymbol{=}\dfrac{h}{\gamma m \upsilon} 
\nonumber  
\end{equation}
  where $\lambda$  is the wavelength, $h$ is the Planck constant, $p$ is the momentum, $m$ is the rest mass, $\upsilon$  is the velocity and $c$ is the speed of light in a vacuum.

Louis de Broglie hypothesis is based on the Special Relativity theory and the associated Lorentz transformations. Note that this picture of a particle with velocity $\upsilon$ and its associated plane phase wave with velocity $\mathrm w=c^2/\upsilon$ is Lorentz invariant, that is the same in any inertial system.

(1)
See my answer here : About de Broglie relations, what exactly is E? Its energy of what?.

