The usual definition of the path integral in QM usually goes as follows:

Let $ [a,b]$ be one interval. Let $ (P_n)$ be the sequence of partitions of $ [a,b]$ given by $ $ P_n=\{t_0,\dots,t_n\}$ $ with $ t_k = t_0 + k\epsilon$ where $ \epsilon = (ba)/n$ , and $ t_0 = a$ , $ t_n=b$ .

Let $ \mathfrak{F}: C_0([a,b];\mathbb{R}^d)\to \mathbb{C}$ be a functional defined on the space of continuous paths on $ [a,b]$ . One defines its discretization as the set of functions $ \mathfrak{F}_n : \mathbb{R}^{(n+1)d}\to \mathbb{C}$ given by $ $ \mathfrak{F}_n(x_0,\dots,x_n)=\mathfrak{F}[\xi_n(t)]$ $ where $ \xi_n(t)$ is the curve defined by taking the partition $ P_n$ , defining $ \xi(t_i)=x_i$ and linearly interpolating between the points – in other words $ \xi(t)$ is for $ t\in [t_i,t_{i+1}]$ the straight line joining $ x_i$ and $ x_{i+1}$ .

One defines the functional integral as the limit $ $ \int_{C_0([a,b];\mathbb{R}^d)}\mathfrak{F}[x(t)]\mathcal{D}x(t)=\lim_{n\to \infty}\int_{\mathbb{R}^{(n+1)d}} \mathfrak{F}_n(x_0,\dots, x_n) d^dx_0\dots d^dx_n$ $
if it exists.

In the case of interest for physics one has $ \mathfrak{F}[x(t)]=e^{iS[x(t)]}$ or rather $ \mathfrak{F}_E[x(t)]=e^{iS_E[x(t)])}$ the euclidean version.
So by slicing the time axis into equal subintervals, one converts the functional to a sequence of functions, integrates those and takes the limit.
This is the construction outlined for instance in Peskin’s book or Sakurai’s book, just rewritten in a more “mathematical” form.
Now, if on the very first step we choose another sequence of partitions $ (P_n)$ such that the sequence of partition’s norms $ P_n\to 0$ as $ n\to \infty$ but which is not the sequence of equal subintervals, would the resulting path integral be different?
I don’t see reason why it should be equal. The intervals endpoints are distinct, the interpolations are distinct, hence the maps $ \mathfrak{F}_n$ are distinct.
If it is I think this is a big problem. After all, the way we are slicing the time axis is arbitrary and just one trick to make the problem easier to deal with.