= Solution
Let $e_0,\ldots,e_{n-1}$ be the standard basis and let the generator $g$ send $e_j$ to $e_{j+1}$ modulo $n$. The constant vector
$$
c_0=\sum_{j=0}^{n-1}e_j
$$
spans a trivial module. Since $n$ is even, the alternating vector
$$
c_{n/2}=\sum_{j=0}^{n-1}(-1)^je_j
$$
spans a sign module on which $g$ acts by $-1$.
For each $1\leq k<n/2$, define the real <discrete Fourier modes>
$$
c_k=\sum_{j=0}^{n-1}\cos\left(\frac{2\pi kj}{n}\right)e_j,
\qquad
s_k=\sum_{j=0}^{n-1}\sin\left(\frac{2\pi kj}{n}\right)e_j.
$$
The plane $W_k=\operatorname{span}\{c_k,s_k\}$ is invariant, and $g$ acts on it by rotation through $2\pi k/n$. Since this angle is neither zero nor $\pi$, $W_k$ is irreducible over $\mathbb R$. The traces $2\cos(2\pi k/n)$ are distinct for $1\leq k<n/2$, so these modules are pairwise nonisomorphic. Orthogonality of the <discrete Fourier transform> gives
$$
\boxed{\mathbb RC_n
=\langle c_0\rangle\oplus\langle c_{n/2}\rangle
\oplus\bigoplus_{k=1}^{(n-2)/2}W_k}.
$$
The dimensions sum to $1+1+2(n-2)/2=n$, so this is the whole regular module: two nonisomorphic one-dimensional summands and $(n-2)/2$ nonisomorphic two-dimensional summands.
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