Solving a (fun!) coequalizer problem for $\mathrm{SL}_n(\mathbb{R})\rightarrow\mathrm{SL}_n(\mathbb{C})$ in $\mathbf{Grp}$

The coequalizer is trivial.

$SL_n(\mathbb{C})$ is almost a simple group (for $n \ge 2$, and it's trivial for $n = 1$): its center $Z(SL_n(\mathbb{C}))$ is the subgroup of scalar multiples of the identity where the scalar is an $n^{th}$ root of unity, and the quotient by the center is the projective special linear group $PSL_n(\mathbb{C})$, which is simple (either as an abstract group or as a Lie group; for simplicity as an abstract group see, for example, this note by Keith Conrad).

This implies that a normal subgroup containing any non-central element of $SL_n(\mathbb{C})$ must in fact be all of $SL_n(\mathbb{C})$, which is certainly the case for the normal subgroup describing this coequalizer.