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Use the four-dimensional real vector space of Hermitian matricesFor $A\in SL(2,\mathbb{C})$, $X\mapsto AX{A}^{†}$ preserves determinant, hence defines a Lorentz transformation. Positive definite $X$ stays positive definite, so future timelike vectors remain future timelike. Connectedness also fixes the orientation, giving an image in $S{O}_0(3,1)$.

If the action is trivial, $X=I$ makes $A$ unitary. The remaining condition says it commutes with every Hermitian matrix, forcing $A=\lambda I$ with ${\lambda }^2=1$. Infinitesimally $aX+X{a}^{†}=0$ first makes $a$ anti-Hermitian and then scalar; tracelessness forces $a=0$. Both real Lie-algebra dimensions are six, so the image is open. Polar decomposition shows $SL(2,\mathbb{C})$ is connected, with connected unitary factor $SU(2)$ and a contractible positive Hermitian determinant-one factor. The image is therefore the full proper orthochronous Lorentz group.This is the Hermitian-matrix double cover of SO0(3,1). The full determinant-one Lorentz group also contains transformations reversing time orientation, so the identity-component qualification cannot be omitted.