The fastest way is to compare kinetic energies in the two cases:
\begin{align*}
KE &= \tfrac{1}{2}I_{\text{cm}}\omega^2_{\text{cm}} + \tfrac{1}{2}M(R\omega)^2_{\text{cm}} \\
KE &=\tfrac{1}{2}I_{\text{inst}}\omega_{\text{inst}}^2 = \tfrac{1}{2} (I_{\text{cm}} + MR^2)\omega^2_{\text{inst}}
\end{align*}
So $\omega_{\text{inst}}=\omega_{\text{cm}}$. The first equation is rotational KE plus translational KE (remember that $\omega = v/R$) for rotation about the center of mass, while the second equation is only rotational energy around the instantaneous axis, where we used the parallel axis theorem.
Note: if the cylinder was sliding then the translational kinetic energy in the first equation will be less than $\tfrac{1}{2}M(R\omega)^2_{\text{cm}} $, so you have to replace the the above system with
\begin{align*}
KE &< \tfrac{1}{2}I_{\text{cm}}\omega^2_{\text{cm}} + \tfrac{1}{2}M(R\omega)^2_{\text{cm}} \\
KE &=\tfrac{1}{2}I_{\text{inst}}\omega_{\text{inst}}^2 = \tfrac{1}{2} (I_{\text{cm}} + MR^2)\omega^2_{\text{inst}}
\end{align*}
there for $\omega_{\text{inst}}<\omega_{\text{cm}}$, consistent with intuition.