In fermion's path integral we have a meassure that you can write, in terms of the Grassman variables $\psi, \bar{\psi}$ as

$$ D\bar{\psi}D\psi, \quad \psi(x) = \sum_n a_n\phi_n(x), \quad \bar{\psi}(x) = \sum_n \bar{a}_n \bar{\phi}_n(x) $$

Where $a_n, \bar{a}_n$ are Grassman variables and $\phi_n(x)$ a set of orthonormal functions such that

$$ \int d^3x\ \phi^\dagger_n(x)\phi_m(y) = \delta_{nm} $$

Now if you perform a change of variables in, for instance, axial group $U(1)_A$ with an small parameter $\alpha(x)$, this renders

$$ a'_m = \sum_n(\delta_{mn} + i\int d^3x\ \alpha(x)\phi^\dagger_m(x)\gamma^5\phi_n(x))a_n = \sum_n(1 + C)_{mn}a_n $$ $$ \bar{a}'_m = \sum_n(1 + C)_{mn}\bar{a}_n $$ $$ C_{mn} = i\int d^3x\ \alpha(x)\phi^\dagger_m(x)\gamma^5\phi_n(x)), \quad 1\ {\rm is\ the\ identity} $$

Now, following Peskin (chapter 19.2, Eq. (19.69)), the path integral meassure should change as

$$ D\bar{\psi}'D\psi' = D\bar{\psi}D\psi·(det[1 + C])^{-2} $$

I don't understand where the -2 power for the jacobian ($det[1 + C]$) came up since if we were talking about a usual integral with usual variables we would end up with +2 power.

What am I missing?

EDIT

Thinking about the problem I found a possible explanation. Grassman's variables, let's call it $\eta$, are forced to satisfy

$$ \int d\eta\ \eta = 1 $$

Therefore, a change of variables such that $\eta$ changes to $$\eta' = A\eta\tag{A}$$ and $\eta'$ is still a Grassman variable should fulfill

$$ \int d\eta'\ \eta' = 1 \tag{B} $$

But if we follow the change of variables given by Eq. (A) and we want Eq. (B) satisfied,

$$ \int d\eta'\ \eta' = A^{-2}\int d\eta\ \eta = 1 \Leftrightarrow \int d\eta'\ \eta' = A^2 $$

Then, we are violating Eq. (B) and $\eta'$ isn't a Grassman variable. So, if $\eta, \eta'$ are Grassman variables then the jacobian ($j = A^{-1}$) among themselves must be introduced in the meassure with the opposite power sign, so:

$$ \int d\eta'\ \eta' = \int j^{-1}·d\eta\ j·\eta \equiv 1 $$

Fine or something to complain about?