A somewhat informal argument, making use of Just and Weese's "architect's view of set theory" is as follows.
Suppose that the collection of all posets isomorphic to $\langle X , \preceq_X \rangle$ were a set, and write this set as $\mathcal{X} = \{ \langle X_i , \preceq_i \rangle : i \in I \}$, where $I$ is some index set. For each $i \in I$ choose an order-isomorphism $f_i : \langle X , \preceq_X \rangle \to \langle X_i , \preceq_i \rangle$.
By $Y$ denote the set $ \{ \langle f_i (x) \rangle_{i \in I} : x \in X \} = \{ \langle x_i \rangle_{i \in I} \in \textstyle{\prod_{i \in I}} X_i : ( \exists x \in X ) ( \forall i \in I ) ( x_i = f_i(x) ) \} .$ We may define a relation $\preceq$ on $Y$ by $\langle f_i (x) \rangle_{i \in I} \preceq \langle f_i (y) \rangle_{i \in I}$ iff $x \leq y$. It is easy to see that $\langle Y , \preceq \rangle$ is order-isomorphic to $\langle X , \leq \rangle$, and therefore $\langle Y , \preceq \rangle \in \mathcal{X}$, i.e., $\langle Y , \preceq \rangle = \langle X_j , \preceq_j \rangle$ for some $j \in I$.
Given any $\langle x_i \rangle_{i \in I} \in Y$ it must be that each $x_i$ was constructed strictly before $\langle x_i \rangle_{i \in I}$, in particular $x_j \in X_j = Y$ was constructed strictly before $\langle x_i \rangle_{i \in I}$. But $x_j = \langle x^{(1)}_i \rangle_{i \in I}$, and so each of these components were constructed strictly before $x_j$; in particular $x^{(1)}_j \in X_j = Y$. etc. etc. ad infinitum.
We then have an infinite sequence of objects/sets, $\langle x_i \rangle_{i \in I} = x^{(0)} , x^{(0)}_j = x^{(1)}, x^{(1)}_j = x^{(2)} , \ldots$ and for each $n$ the object/set $x^{(n+1)}$ must be constructed strictly before $x^{(n)}$. But this means that at no point were any of the $x^{(n)}$ ready to be constructed!