A total order $I$ is said to be weakly self-similar if there exists a proper subset $J \subsetneq I$ together with a bijective, order-preserving function $f:I \to J$ (that is, $J$ is isomorphic to $I$). Analogously, $I$ is said to be strongly self-similar if there exists an element $i \in I$ such that one of the total orders $$I_{< i}, I_{\le i}, I_{> i}, I_{\ge i}$$ is isomorphic to $I$. Such sets are defined as you imagine, e.g. $I_{< i} = \{ x \in I: x < i \}$.
Does there exist self-similar total orders on a countable set, whether in a weak or strong sense?
As noted in the comments, the question is quite trivial: $\omega$ provides an example for both questions. Regarding a more stringent definition, that is $I_{<i}$ and $I_{\ge i}$ being both isomorphic to $I$, one can consider $\mathbb{Q} \cap [0,1)$ and splitting at any value $q \in \mathbb{Q} \cap (0,1)$, e.g. $q = 1/2$.
As a side note, observe that such countable total orders cannot define a "fractal behavior": if we define $T^0 I := I_{< i}, T^1 I := I_{\ge i}$ as the two orders in which $I$ is split, 'most' of the sets $$ T^{\textbf{c}} I := \bigcap_{n=1}^{\infty} T^{c_n} \ldots T^{c_1} I$$ for $\textbf{c} := (c_1, \ldots, c_n, \ldots ) \in \{0,1\}^{\mathbb{N}}$ will be empty, since there are uncountably many such intersections. This is what misled me...