We can say something about the sizes of the antichains in infinite posets $P$ satisfying $P\cong \textrm{End}(P)$.
Lemma. If $P$ is an infinite poset, then
End($P$) has an infinite antichain, and
if $P$ has an infinite antichain of cardinality $\kappa$, then End($P$) has an antichain of cardinality $2^{\kappa}$.
Sketch of proof of item 1. Case 1: $P$ has an infinite antichain. Then End($P$) must also have one, since there is an embedding of $P$ into End($P$) which takes $x$ to the constant function with range $\{x\}$. Case 2: $P$ has no infinite antichain. In this case $P$ must contain an increasing or decreasing $\omega$-sequence. From this it can be argued that $P$ must have an endomorphism $\sigma\colon P\to P$ whose range $\sigma(P)$ is either an increasing $\omega$-sequence or an increasing $\omega+1$-sequence.
Here is how one produces such a $\sigma$. Suppose that $\Sigma: p_0<p_1<\cdots$ is an increasing $\omega$-sequence. Case 1: $\Sigma$ is cofinal in $P$. In this case, map $x\in P$ to the least $p_i$ such that $x\not\geq p_i$ Case 2: $\Sigma$ is not cofinal, because there is an element $q$ above every $p_i$. In this case, map $x\in P$ to the least $p_i$ such that $x\not\geq p_i$ if such a $p_i$ exists. If $x\geq p_i$ for every $i$, map $x$ to $q$. In Case 1, the range of $\sigma$ is an $\omega$-sequence, while in Case 2 it is an $\omega+1$-sequence.
It is not hard to show that End($\omega$) and End($\omega+1$) each have infinite antichains, so End($\sigma(P)$) has an infinite antichain. If $\{\alpha_0, \alpha_1, \ldots\}$ is an infinite antichain in End($\sigma(P)$), then $\{\alpha_0\circ\sigma, \alpha_1\circ\sigma, \ldots\}$ is an infinite antichain in End($P$).
Sketch of proof of item 2. Case 1: $P$ is an antichain. In this case, End($P$) is an antichain of cardinality $2^{|P|}$, so there is really nothing to do. Case 2: $P$ contains elements $a<b$. Let $A\subseteq P$ be an antichain of cardinality $\kappa$. It is possible to find a set $\mathcal X$ of pairwise incomparable subsets of $A$ such that $|{\mathcal X}|=2^{\kappa}$. For each $U\in{\mathcal X}$ define $f_U\colon P\to P$ by $f_U(x)=b$ if $x$ is in the filter generated by $U$ in $P$, else $f_U(x)=a$. The set $\{f_U | U\in\mathcal X\}$ is an antichain of size $2^{\kappa}$ in End($P$). \\
So if $P$ is an infinite poset satisfying $P\cong \textrm{End}(P)$, then it has antichains of size $\beth_n$ for every $n$.