This isn't true. I follow Paul's suggestion of taking (discrete) Fourier transforms $\widehat{x}(\zeta)=\sum_{j=0}^{n-1} x_j\zeta^j$, $\zeta^n=1$. Then a subspace closed under convolution becomes one closed under pointwise multiplication, and $\widehat{\pi x} =\zeta^{-1}\widehat{x}$.
Now the subspace defined by the condition that $\widehat{x}(\zeta_1)=\widehat{x}(\zeta_2)$ is closed under multiplication, but won't be left invariant by multiplication by $\zeta$. We can make this more concrete, also to confirm that we can obtain real valued examples: Take $n=4$ and the subspace generated by (taking linear combinations of convolution powers of) $x=(1,-1,1,1)$. Notice that $\widehat{x}(1)=\widehat{x}(-1)=2$.
Alternatively, we can forget how we got this and just check directly that the subspace generated by $x$ is spanned by $e_1,x,e_3$, and this is not invariant under $\pi$.