Let's discretize time $t$. In other words, assume that we have $n$ Grassmann-odd variables $\theta^1, \ldots, \theta^n$. In this language OP's differential operator $\partial_t$ is replaced with some Grassmann-even matrix $A^i{}_j$. Define $$\theta^{\prime i}= \sum_{j=1}^nA^i{}_j \theta^j.$$

Then a straightforward calculation shows that OP's path-integral is replaced with $$\begin{align} \int d\theta^n \ldots d\theta^1~\theta^{\prime 1}\ldots \theta^{\prime n}~=~&\ldots~=~ \int d\theta^n \ldots d\theta^1~\delta(\theta^{\prime 1})\ldots \delta(\theta^{\prime n})\cr ~=~&\ldots~=~\det(A),\end{align} $$ which indeed is a discrete version of OP's claim (2).

TL;DR: Yes, OP is correct.

Let's discretize time $t$. In other words, assume that we have $n$ Grassmann-odd variables $\theta^1, \ldots, \theta^n$. In this language OP's differential operator $\partial_t$ is replaced with some Grassmann-even matrix $A^i{}_j$. Define $$\theta^{\prime i}= \sum_{j=1}^nA^i{}_j \theta^j,$$ so that $$\frac{\partial}{\partial\theta^j} ~=~\sum_{i=1}^n A^i{}_j\frac{\partial}{\partial\theta^{\prime i}}.$$ Then a straightforward calculation shows that OP's path-integral (1) is replaced with$^1$ $$\begin{align} \int\!d\theta^n \ldots d\theta^1~\theta^{\prime 1}\ldots \theta^{\prime n}~=~&\ldots~=~ \int\! d\theta^n \ldots d\theta^1~\delta(\theta^{\prime 1})\ldots \delta(\theta^{\prime n})\cr ~=~&\ldots~=~\det(A),\end{align} $$ which indeed is a discrete version of OP's claim (2).

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$^1$Recall that Berezin integration $$ \int\!d\theta^n \ldots d\theta^1 ~=~\frac{\partial}{\partial\theta^n}\ldots \frac{\partial}{\partial\theta^1}$$ is the same as differentiation!

Let's discretize time $t$. In other words, assume that we have $n$ Grassmann-odd variables $\theta^1, \ldots, \theta^n$. In this language OP's differential operator $\partial_t$ is replaced with some Grassmann-even matrix $A^i{}_j$. Define $$\theta^{\prime i}= \sum_{j=1}^nA^i{}_j \theta^j.$$

Then a straightforward calculation shows that OP's path-integral is replaced with $$\begin{align} \int d\theta_n \ldots d\theta_1~\theta^{\prime 1}\ldots \theta^{\prime n}~=~&\ldots~=~ \int d\theta_n \ldots d\theta_1~\delta(\theta^{\prime 1})\ldots \delta(\theta^{\prime n})\cr ~=~&\ldots~=~\det(A),\end{align} $$ which indeed is a discrete version of OP's claim (2).

Let's discretize time $t$. In other words, assume that we have $n$ Grassmann-odd variables $\theta^1, \ldots, \theta^n$. In this language OP's differential operator $\partial_t$ is replaced with some Grassmann-even matrix $A^i{}_j$. Define $$\theta^{\prime i}= \sum_{j=1}^nA^i{}_j \theta^j.$$

Then a straightforward calculation shows that OP's path-integral is replaced with $$\begin{align} \int d\theta^n \ldots d\theta^1~\theta^{\prime 1}\ldots \theta^{\prime n}~=~&\ldots~=~ \int d\theta^n \ldots d\theta^1~\delta(\theta^{\prime 1})\ldots \delta(\theta^{\prime n})\cr ~=~&\ldots~=~\det(A),\end{align} $$ which indeed is a discrete version of OP's claim (2).