Initially n identical capacitors are joined in parallel and are charged to potential V. Now they are separated and joined in series. Then

The initial parallel combination of $n$ identical capacitors each with capacitance $C$ gives equivalent capacitance $C_p = nC$, charged to potential $V$. The stored energy is $U_p = \frac{1}{2}nCV^2$, and each capacitor carries charge $CV$.

When reconfigured in series, the equivalent capacitance becomes $C_s = C/n$. Each capacitor retains its charge $CV$, so the series combination carries total charge $Q_s = CV$.

The resulting potential difference is $V_s = Q_s / C_s = CV / (C/n) = nV$. The energy stored is $U_s = \frac{1}{2}C_s V_s^2 = \frac{1}{2}(C/n)(nV)^2 = \frac{1}{2}nCV^2$.

The potential difference increases by a factor of $n$ while the total energy remains unchanged.

Final answer: $\boxed{C}$