Choosing a tree (t_2) that's a line (that is, (N) nodes connected as a chain) always minimizes (\text{similarity}(S(t_1), S(t_2))).
Proof: Let (t_\text{line}) be a line of size-(N), and (t_x) be any tree of size (N). We'll show for any (t_1) and any (t_x), (\text{similarity}(S(t_1), S(t_\text{line})) \le \text{similarity}(S(t_1), S(t_x)))). Let’s also not consider subtrees of size (1) (because all trees of size (n) have (n) subtrees of size (1)). There are two cases:
Case 1: Consider (t_x) such that (\text{diameter}(t_x) \le \text{diameter}(t_1)). We can choose any two nodes from (t_x) and get the path between them. This will be a subtree of (t_1). However, (\text{similarity}(S(t_1), S(t_\text{line}))) is at most (\binom{n}{2}), because (S(t_\text{line})) itself has only (\binom{n}{2}) elements.
Case 2: Consider (t_x) such that (\text{diameter}(t_x) > \text{diameter}(t_1)). For any pair of nodes ((u, v) \in t_x), either the path from (u) to (v) is a subtree of (t_1), or there is a corresponding pair in (t_\text{line}) that is not a subtree of (t_1).
Therefore, (\text{similarity}(S(t_1), S(t_\text{line})) \le \text{similarity}(S(t_1), S(t_x))). To compute the former, simply calculate the diameter of (t_1) using two DFS's and use combinatorics. In particular, each length (i) line (of (i + 1) nodes) appears in (S(t_\text{line})) exactly (N - i) times, and all of these sub-lines are isomorphic to a subtree of (t_1). The answer will be: [ \sum_{i=0}^{\text{diameter}(t_\text{line})} N - i ]