Let $E$ be a separable $\mathbb R$-Banach space, $\rho$ be a complete separable metric on $E$, $\operatorname W\_\rho$ denote the Wasserstein metric of order $1$ associated to $\rho$, $\mathcal M\_1(E)$ denote the set of probability measures on $(E,\mathcal B(E))$ and $(\kappa\_t)\_{t\ge0}$ be a Markov semigroup on $(E,\mathcal B(E))$ with $$\operatorname W\_\rho(\mu\kappa\_t,\nu\kappa\_t)\le ce^{-\lambda t}\operatorname W\_\rho(\mu,\nu)\;\;\;\text{for all }\mu,\nu\in\mathcal M\_1(\mu,\nu)\tag1$$ for some $c\ge0$ and $\lambda>0$.



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> Are we able to conclude that $(\kappa\_t)\_{t\ge0}$ has a unique invariant measure $\mu\_\ast\in\mathcal M\_1(E)$?
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By $(1)$, the adjoint semigroup $(\kappa\_t^\ast)\_{t\ge0}$ is eventually contractive: Let $t\_0\ge0$ with $$ce^{-\lambda t}<1\;\;\;\text{for all }t\ge t\_0$$ and $t\ge t\_0$. Since the Wasserstein space $$\mathcal S^1(E,\rho):=\left\{\mu\in\mathcal M\_1(E):(\mu\otimes\delta\_0)\rho<\infty\right\}$$ equipped with $\operatorname W\_\rho$ is complete and hence we can apply Banach's fixed-point theorem yielding that there is a unique $\mu\_\ast\in\mathcal S^1(E,\rho)$ with $$\mu\_\ast\kappa\_t=\mu\_\ast\tag2.$$ Moreover, for any $\mu\_0\in\mathcal S^1(E,\rho)$ and $$\mu\_n:=\mu\_{n-1}\kappa\_t\;\;\;\text{for }n\in\mathbb N,$$ it holds $$\operatorname W\_\rho(\mu\_n,\mu\_\ast)\xrightarrow{n\to\infty}0\tag3.$$



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> So, all what's left to prove is that $\mu\_\ast$ does not depend on $t$, i.e. $\mu\_\ast$ is invariant with respect to $\kappa\_t$ for all $t\ge t\_0$.
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BTW: Is this all we can hope for or can we even conclude that $\mu\_ast$ must be invariant with respect to $\kappa\_t$ for all $t\ge\color{red}0$?



If a Markov semigroup is eventually contractive, can we conclude that it admits a unique invariant measure?
Note that your argument contains an implicit assumption that $\kappa\_t \mu \in \mathcal{S}^1$ for every $\mu \in \mathcal{S}^1$ (otherwise the Banach fixed point theorem does not apply). I will also make that assumption. Also, I realized that I have written $\kappa\_t \mu$ with $\mu$ on the right; sorry about that.


You have shown that for some fixed $t^\* \ge t\_0$, that $\kappa\_{t^\*}$ has an invariant measure $\mu\_\*$ which is unique in $\mathcal{S}^1$.


Let $t > 0$ be arbitrary. Then we have by the semigroup property that
$$\kappa\_{t^\*} \kappa\_t \mu\_\* = \kappa\_{t+ t^\*} \mu\_\* = \kappa\_t \kappa\_{t^\*} \mu\_\* = \kappa\_t \mu\_\*$$
which proves that $\kappa\_t \mu\_\*$ is invariant for $\kappa\_{t^\*}$. By uniqueness, $\kappa\_t \mu\_\* = \mu\_\*$. This proves that $\mu\_\*$ is invariant for $\kappa\_t$.


If $t \ge t\_0$, then your argument shows that $\mu\_\*$ is in fact the unique invariant measure in $\mathcal{S}^1$ for $\kappa\_t$. Otherwise, for $t < t\_0$, suppose $\mu' \in \mathcal{S}^1$ is another invariant measure for $\kappa\_t$. Let $n$ a large enough integer so that $n t \ge t\_0$; then $\mu' = \kappa\_t^n \mu' = \kappa\_{nt} \mu'$. Since $\kappa\_{nt}$ has $\mu\_\*$ as its unique invariant measure, we have $\mu' = \mu\_\*$.


We have thus shown that for every $t$, $\mu\_\*$ is invariant for $\kappa\_t$, and is the unique such measure in $\mathcal{S}^1$.


