I've had to rewrite the question as i made agrievious errors in the first go-round.
Given a (pseudo)Riemannian metric g, we can identify its components with the symmetric product of gamma matrices:
$$\gamma^{\mu}\gamma^{\nu}+\gamma^{\nu}\gamma^{\mu}=2g^{\mu\nu}I$$
The gammas act as basis vectors and their products and linear combinations form a clifford algebra. Under the geometric algebra our basis vectors (matrices now) transform in a two-sided way:
$$\bar{\gamma}^{\mu}=\psi^{\dagger}\gamma^{\mu}\psi$$
for a general non-unitary transformation (i,e not just a change of coordinate basis) $\psi^{\dagger}\neq\psi^{-1}$ how does the metric g and its determinant Det(g) transform??
I had initially thought $g\longrightarrow\psi^{\dagger}g\psi$ however that is silly since we have the non-unitary generalization.
$$g^{\mu\nu}I=\frac{1}{2}\psi^{\dagger}\left(\gamma^{\mu}\psi\psi^{\dagger}\gamma^{\nu}+\gamma^{\nu}\psi\psi^{\dagger}\gamma^{\mu}\right)\psi$$
For my problem set $\psi\psi^{\dagger}$ should be a scalar (in case you havent guessed the $\psi$ are spinors). this should leave me with:
$$g^{\mu\nu}I=\frac{1}{2}\psi^{\dagger}\left(\gamma^{\mu}\gamma^{\nu}+\gamma^{\nu}\gamma^{\mu}\right)\psi\left(\psi\psi^{\dagger}\right)$$