Usual distances on DFAs (Deterministic Finite Automata)?
There are many possible distance metrics, and without any criteria, we have no basis to choose.  Here are two plausible ones.

Let $L_1,L_2$ be the languages of the two DFAs.  Let $L$ be the symmetric difference of those languages (i.e., $(L_1 \setminus L_2) \cup (L_2 \setminus L_1)$).

One dissimilarity measure is $d(L_1,L_2) = 2^{-n}$ where $n=\min\{|x| : x \in L\}$, i.e., $n$ is the length of the shortest word in $L$.  This is not a distance metric, because it does not satisfy the triangle inequality.

Another possibility is to use the density of $L$, i.e.,

$$d(L_1,L_2) = \sum_{n=0}^\infty {|L \cap \Sigma^n| \over 2^n |\Sigma|^n}.$$

This one is a distance metric.

These can be computed given DFAs for $L_1,L_2$.  You can easily compute a DFA for $L$.  Then, it is easy to find the length of the shortest word in $L$ using breadth-first search.

Computing the density is a bit trickier, but can be done efficiently using a little more math.  Let $A$ denote the transition matrix of the DFA for $L$, i.e., $A_{ij}=1$ if there is a transition (on any input symbol) from state $i$ to state $j$, or $A_{ij}=0$ if not.  Let $s$ denote the one-hot vector indicating the start state, i.e., $s_i=1$ iff $i$ is the start state, and $f$ denote the one-hot vector indicating the final states, i.e., $f_i=1$ iff $i$ is an accepting state.  Now $|L \cap \Sigma^n| = s^\top A^n f$, so the distance is given by

$$d(L_1,L_2) = \sum_{n=0}^\infty {s^\top A^n f \over 2^n |\Sigma|^n}
= s^\top C f$$

where

$$C = \sum_{n=0}^\infty B^n = {1 \over I - B} = (I - B)^{-1}$$

where $B={1 \over 2|\Sigma} A$ and $I$ represents the identity matrix.  This can be computed in time proportional to the cube of the number of states in the DFA for $L$ using standard algorithms for matrix inversion.