Consider the following Itō diffusion: $${\rm d}X_t=b(X_t){\rm d}t+\sigma(X_t){\rm d}W_t\tag1$$ and the corresponding Euler-Maruyama discretization $$Y_i:=Y_{i-1}+\Delta tb(Y_{i-1})+\sqrt{\Delta t}\sigma(Y_{i-i})\xi_i\tag2,$$ where $\Delta t>0$ and $(\xi_i)_{i\in\mathbb N}$ is an independent identically $\mathcal N_{I_d}(0,\;\cdot\;)$-distributed process.
Note that $(Y_i)_{i\in\mathbb N_0}$ is a Markov cain with transition kernel $$\kappa(x,B):=\mathcal N_{\Delta t\sigma(x)^\ast\sigma(x)}(x+\Delta tb(x),B)\;\;\;\text{for }(x,B)\in\mathbb R^d\times\mathcal B(\mathbb R^d)\tag3.$$
Question: Are we able to give an explicit formula for an $\kappa$-invariant measure $\mu$? (Possibly under specific assumptions on $b,\sigma$ and/or $\Delta t$.)
Let's first try to prove that this is not possible to see if we need to give a negative answer. For simplicity, assume $d=1$. Assume $\mu$ exists. Then, $$\mu((-n,n))\le2n\int\mu({\rm d}x)\frac1{\sqrt{2\pi\Delta t\sigma^2(x)}}\tag4$$ for all $n\in\mathbb N$. From this we immediately see that when $\sigma$ is constant, then $\mu$ cannot exists (since then the right-hand side of $(4)$ tends to $0$ as $t\to\infty$; contradicting that $\mu$ is a probability measure.)
So, we definitely need to assume $\sigma$ is not constant ... In that case, we don't arrive immediately at a contradiction. So, can we proceed and give a positive answer?
EDIT
Maybe I'm wrong, but when $\mu$ exists, it should be a normal distribution. So, it would be enough to figure out the mean and variance of it. Now, for the mean: $$m:=\int\mu({\rm d}x)x=\int\mu({\rm d}x)\int\kappa(x,{\rm d}y)y=m+\Delta t\int b\:{\rm d}\mu\tag5;$$ Unfortunately, this is not really useful in determining $m$; but it seems to require $$\int b\:{\rm d}\mu=0\tag6.$$ Now, if $b$ would be nonnegative, this seems to enforce $b=0$ (since it enforces $b\varphi=0$ Lebesgue almost everywhere, where $\varphi$ is the density of the normal distribution which we are assuming $\mu$ is; since that density is strictly positive, we need to have $b=0$ Lebesgue almost everywhere). So, it seems like we need that $b$ is neither nonnegative, nor nonpositive.
Assuming, for simplicity, $d=1$, we also obtain $$\varsigma^2:=\int\mu({\rm d}x)(x-m)^2=\Delta t^2\int b^2\:{\rm d}\mu+\Delta t\int\sigma^2\:{\rm d}\mu+2\Delta t\int\mu({\rm d}x)xb(x)+\varsigma^2\tag7.$$