circuit analysis - Order of a system

Does the order of a system always depend on the number of independent initial conditions?

Consider this single mesh containing a DC voltage source, one resistor, and two capacitances and a switch which closes at t = 0. Nothing is stopping me from having two different initial voltages on these two caps and satisfying KVL all the time.

^{simulate this circuit – Schematic created using CircuitLab}

$$V_{c1}\:\textrm{and}\:V_{c2} \textrm{ are initial voltages on the caps.} $$

Now, I also know that since these two caps are in series writing a KVL around the mesh will give an equation with a single equivalent capacitor and hence deducing that order of the system is one.

Which can also be verified by writing its transfer function. (Obviously when there are no initial conditions.)

\begin{equation} \frac{V_o(s)}{V_i(s)} = \frac{C_1}{{C_1}+{C_2}} \frac{1 + s{C_2}R}{1 + s{C_{eq}}R} \end{equation}

where $$C_{eq} = \frac{{C_1}{C_2}}{{C_1} + {C_2}}$$

Clearly, there is only one pole and hence order is one.

This confuses me. Is this some degenerate case or the statement that the order of a system is equal to the number of independent initial conditions is somewhat flawed.

EDIT 1

Now if I take this circuit.

^{simulate this circuit}

As far as the structure of the network is concerned nothing has changed since the input ideal current source can be taken as an open circuit and I know this for a fact that injection of the input and the pair of terminals across which we're taking the output doesn't alter the location of the poles as long as the structure remains invariant.

This circuit has the same structure as that of the earlier one.

But the order here is two in agreement with the statement that order of a system is equal to the number of independent initial conditions as shown below.

\begin{equation} \frac{V(s)}{I(s)} = \frac{1}{s(C_1 + C_2)} \frac{1+sC_{2}R}{1+sC_{eq}R} \end{equation}

These two examples are countering each other.