This seems like it should be simple but somehow I do not see how.
The Majorana Lagrangian can be written in terms of a left handed Weyl spinor $\psi_L$ as
$$ \mathcal{L}_M= i \psi_L^\dagger \bar{\sigma}^\mu \partial_\mu \psi_L - \tfrac{m}{2} \psi^T_L \epsilon \psi_L + \tfrac{m}{2} \psi^\dagger_L \epsilon \overline{\psi}_L \hspace{1cm}. $$ Meanwhile the Dirac Lagrangian can be written in terms of a Left handed Weyl spinor $\psi_L$ and a right handed Weyl spinor $\psi_R$ as $$ \mathcal{L}_D = i \psi_L^\dagger \bar{\sigma}^\mu \partial_\mu \psi_L + i \psi_R^\dagger \sigma^\mu \partial_\mu \psi_R - m (\psi_L^\dagger \psi_R + \psi_R^\dagger \psi_L). $$
Here I am using the convention $\sigma^\mu = (I, \sigma^i)$, $\bar\sigma^\mu = (I, -\sigma^i)$, and $\epsilon = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$.
The reality condition for the Majorana spinor is just $$ \psi_R = - \epsilon \overline{\psi}_L. $$ Plugging in the above $\psi_R$ into $\mathcal{L}_D$, and using the identity $- \epsilon \sigma^\mu \epsilon = (\bar\sigma^\mu)^*$, I get
$$ \mathcal{L}_D = i \psi_L^\dagger \bar{\sigma}^\mu \partial_\mu \psi_L + i \psi_L^T (\bar{\sigma}^\mu)^* \partial_\mu \overline{\psi}_L + m \psi_L^\dagger \epsilon \overline{\psi}_L - m \psi_L^T \epsilon \psi_L. $$
It seems to me that I would have $\mathcal{L}_D = 2 \mathcal{L}_M$ if only I could prove $$ i \psi_L^\dagger \bar{\sigma}^\mu \partial_\mu \psi_L \stackrel{?}{=} i \psi_L^T (\bar{\sigma}^\mu)^* \partial_\mu \overline{\psi}_L. $$
However, I do not see why the above equation has to be true. An added layer of complication is that $\psi_L$ is really a vector of Grassmann variables that satisfy $$ \{ \psi_L^a, \psi_L^b \}_+ = 0 \hspace{1 cm} \{ \overline{\psi^a}_L, \overline{\psi^b}_L \}_+ = 0 \hspace{1 cm} \{ \psi_L^a, \overline{\psi^b}_L \}_+ = \delta^{ab} $$ for $a,b = 1,2$.
What are the correct manipulations to show that $\mathcal{L}_D = 2 \mathcal{L}_M$?