Looking at the following Lagrangian for a non-Abelian gauge theory with gauge fields $A^a_{\mu}$, scalar matter fields $\phi_i$ with Faddeev-Popov ghosts $c^a$ and anti-ghosts $\bar{c}^a$ included

\[\mathcal L_{FP} = \mathcal L [A^a_{\mu},\phi_i] - \frac{1}{2\xi}(\partial_{\mu}A^a_{\mu})^2 + (\partial_{\mu}\bar{c}^a )(D_{\mu}c^a)\]

with

\[D_{\mu}c^a = \partial_{\mu}c^a+gf^{abc}A^b_{\mu}c^c\]

the BRST transformation under which this Lagrangian is invariant is given by

$$ \phi_i \rightarrow\phi_i + i\theta c^a T^a_{ij}\phi_j$$

$$ A^a_{\mu} \rightarrow A^a_{\mu} + \frac{1}{g}\theta D_{\mu}c^a$$

$$ \bar{c}^a \rightarrow \bar{c}^a - \frac{1}{g}\theta\frac{1}{\xi}\partial_{\mu} A^a_{\mu} $$

$$ c^a \rightarrow c^a - \frac{1}{2}\theta f^{abc}c^b c^c $$

where $\theta$ is a Grassmann number.

I have heard that (informally speaking?) such a BRST transformation can be considered to be half a supersymmetry transformation, probably because the contributions of loops containing the fermionic ghost fields are able to cancel the effects of certain bosonic loops which makes the theory renormalizble (?).

How can I explicitly see that the BRST transformation is half a supersymmetry transformation (if it really is), or what is the exact relationship between the BRST and (for example $\mathcal N = 1$) supersymmetry?

Using the superspace formalism, ($\mathcal N = 1$) supersymmetry can be considered to be a translation in two additional fermionic coordinates, so could a BRST transformation be written down as a translation in a single additional fermionic direction?