Let $X_t$ be a continuous time Markov process on $\mathbb{R}^d$ (or $\mathbb{R}$) with transition kernel given by $$\mathbb{P}[X_t \in dy | X_0 = x] = p_t(x,dy) = \frac{1}{t^{d/2}c} 1_{B_{\sqrt{t}}(x)}(y) dy$$ where $c=\textrm{Leb}(B_1^d)$ is the volume of the unit ball in $\mathbb{R}^d$.
What will such a process look like? Is it even well-defined? Does it relate to Brownian Motion?
The motivation for this is a discussion about using diffusion on an open subset $U \subset \mathbb{R}^d$ to smoothen out a boundary $\partial \Omega \subset U$. There the idea arised to instead of taking a Brownian motion and its corresponding heat kernel, take a uniform kernel on a ball of radius $\sqrt{t}$ corresponding to the above mentioned Markov process.
My thoughts so far:
Since $p_{t+s} \neq p_t \ast p_s$ (indicator functions are not stable under convolution), $p_t$ does not form a probability semigroup. Hence, $X_t$ is not a time-homogenous (as fKonrad corrected) process and can neither have a generator nor solve a classical SDE.
Its paths are probably continuous. This I can only guess from the transition kernel, but if we assume independent and stationary increments (which I did not manage to prove either and doubt due to the previous point), we can apply Kolmogorov's continuity criterion: For every $\alpha, \beta:=\alpha+d-1 > 0$ and $t>s>0$, we have $\mathbb{E}^x[| X_t - X_s | ^{\alpha} ] = \mathbb{E}^0[| X_{t-s} | ^{\alpha} ] = \int \limits_{B_{t-s}(0)} |x|^{\alpha} dx = \int \limits_0^{t-s} \int \limits_{\partial B_{r}(0)} r^{\alpha} dS dr = \int \limits_0^{t-s} cd r^{\alpha +d-1} dr = \frac{cd}{{\alpha}+d}(t-s)^{\alpha+d}$.
Since the transition kernel is radially symmetric and only shifted for different locations $x$, this process should be similar to a Brownian Motion.
You have already shown that the transition functions do not satisfy the Chapman-Kolmogorov equation $p_{t+s}=p_tp_s$. Hence it cannot be a time-homogeneous Markov process.
As you already noted, usually you show existence of a continuous version of $X$ by Kolmogorov's continuity criterion. However, assuming $X$ has stationary and independent increments is not correct, as $X$ would have to be a Lévy process.
I don't think one can meaningfully use Kolmogorov's theorem here, for the following reason:
Let $ s\leq t$ then \begin{align*}\mathbb E[ |X_s-X_t|^\alpha]&=\int_{\mathbb R^d}\mathbb E\bigg[|X_t-X_s|^\alpha\bigg| X_s=x\bigg]\mathbb P^{X_s}(dx)=\int_{\mathbb R^d}\int_{B_{\sqrt {t-s}}(x)}|y-x|^\alpha \frac{1}{(t-s)^{d/2}c}dy\mathbb P^{X_s}(dx)\\ &= \frac{1}{(t-s)^{d/2}c} \int_{\mathbb R^d}\int_{B_{\sqrt {t-s}}(x)}|y-x|^\alpha dy\mathbb P^{X_s}(dx) \end{align*} As you can see the inner integral fails to converge for all $\alpha>d$, hence no estimate satisfying Kolmogorov's continuity condition can be made in this case.
On the other hand, choosing $\alpha<d$ makes the integral converge. In this case since $|y-x|^\alpha$ is radially symmetric around $x$ we get $$\int_{B_{\sqrt{t-s}}(x)}|y-x|^\alpha dy = d\cdot \lambda(B_{\sqrt{t-s}}(0))\cdot \int_0^{\sqrt{t-s}} z^\alpha dz = d\cdot \lambda(B_{\sqrt{t-s}}(0))\cdot \bigg[ \frac{1}{\alpha+1}z^{\alpha +1}\bigg]_0^{\sqrt{t-s}}=d\cdot \lambda(B_{\sqrt{t-s}}(0))\cdot \frac{1}{\alpha+1} |t-s|^{(\alpha+1)/2}$$ I hope i didn't make any mistakes.