Identical one-time marginals do not determine a Markov chain

Let $(X_n)$ be IID with [F1] $\mathbb{P}(X_n=0)=\mathbb{P}(X_n=1)=1\mathrm{/}2$. By [F1], it is Markov with $K(x,\{0\})=K(x,\{1\})=\frac{1}{2}$ for both $x$. Independence gives $\mathbb{P}(X_0=X_1)=\mathbb{P}(0,0)+\mathbb{P}(1,1)=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\mathrm{.}$

Let $B$ be one fair bit and put $Y_n=B$ for every $n$. Then each $Y_n$ is [F2] again fair, while [F2] makes $Y$ Markov with identity kernel $J$. Here $\mathbb{P}(Y_0=Y_1)=1.$ Moreover $J(0,\{0\})=1\mathrm{\ne }1\mathrm{/}2=K(0,\{0\})$, so the kernels differ.

Thus $\mathcal{L}(X_n)=\mathcal{L}(Y_n)$ for every $n$, including $n=0$, [step 1.1, step 1.2] but their displayed two-time event probabilities are $1\mathrm{/}2$ and $1$. This is a concrete failed conclusion: one-time marginals do not determine even a two-time law, much less the kernel or path law. The state space is nonempty and finite; probabilities zero and one appear explicitly. Choice is inherited only from the Markov-chain interfaces used in [F1]--[F2].