Suppose that $D\subset\mathbb{R}^m$ and $g(\cdot)$ is a smooth function mapping $D$ into $\mathbb{R}$ with a unique minimum at $\hat{x}$ lying in the interior of $D$. Then, the Laplace's approximation to $$ I=\int_De^{-g(x)}dx\tag{*} $$ is $$ \frac{(2\pi)^{m/2}}{|g''(\hat{x})|^{1/2}}\exp(-g(\hat{x})). $$ Claim: the Laplace's approximation is invariant to nonsingular linear transformation of the variable of integration.
Edit 2: I will put my own answer below and leave it there just in case it may be useful for someone in the future.
Suppose $x=\phi(y)=Ay+b$ where $A$ is nonsingular, then the substitution rule gives $$ I=\int_De^{-g(x)}dx=\int_{\phi^{-1}(D)}e^{-g(\phi(y))}|\det D\phi(y)|dy\\ =\int_{\phi^{-1}(D)}e^{-g(\phi(y))}|\det A|dy\\ =\int_{\phi^{-1}(D)}\exp[-(g(\phi(y))-\log|\det A|)]dy. $$ If $g(x)$ is minimized at $\hat{x}$, then $g(\phi(y))$ is minimized at $\hat{y}=\phi^{-1}(\hat{x})$. Letting $h(y)=g(\phi(y))-\log|\det A|$, we can apply the Laplace's approximation to the last integral above to get $$ \frac{(2\pi)^m}{|h''(\hat{y})|^{1/2}}\exp(-h(\hat{y}))=\frac{(2\pi)^m}{|h''(\hat{y})|^{1/2}}\exp(-g(\hat{x}))|\det A|. $$ Now the proof is done if we can show that $$ \frac{1}{|g''(\hat{x})|^{1/2}}=\frac{|\det A|}{|h''(\hat{y})|^{1/2}}. $$ But this last step is just the chain rule: $$ h''(\hat{y})=g''(\phi(\hat{y}))(A')^2=g''(\hat{x})(A')^2\implies \det(h''(\hat{y}))=\det(g''(\hat{x}))(\det A)^2. $$