If I understand your approach correctly, it is greedy algorithm, i.e. it is fast, albeit not necessarily finding the optimal solution.
The point is that C_2i could have more customers in common with a cluster other than C_11, i.e. could be a better match for a different cluster C_1i.
That is why I suggest the following:
A clustering $C$ assigns two each customer $u_i, i=1,\ldots k$ one of five labels:
$$
C: u \to \{1,2,\ldots,5\}.
$$
And for a given clustering $C_t$ there are thus $5! = 120$ different clusterings $C_{tr}$ that are mere "label re-assignments", i.e. that lead to the same grouping, with other words such that:
$$
\forall_{ij} \quad C_{tr}(u_i) = C_{tr}(u_j) \Leftrightarrow C_t(u_i) = C_t(u_j).
$$
Now consider two consecutive clusterings $C_t, C_{t+1}$. Fix the label assignment of $C_t$ and run through all possible 120 cluster re-labelings $C_{(t+1)r}$ of $C_{t+1}$ and each time record the count $s_r$ of how many customers have the same cluster label in both clusterings:
$$
s_r = \#\{i| C_{(t+1)r}(u_i) = C_t(u_i)\}.
$$
Then choose the labeling $C_{(t+1)r_o}$ that gives the highest value $s_r$:
$$
r_o = \operatorname{argmax}_r \{s_r\}.
$$
Then, the $q$-th cluster of $C_{(t+1)r_o}$ represents the time evolution of the $q$-th cluster of $C_t$, $q=1,\ldots,5$.