Eigenspace of Gaussian Markov operator Consider the (one-dimensional) Gaussian distribution $Q := N(\nu,\tau^2)$ and the (Gaussian) Markov operator
\begin{equation*}
\begin{array}{rccc}
R : & L_1(\mathbb{R},\mathcal{B}(\mathbb{R}),Q) & \to & L_1(\mathbb{R},\mathcal{B}(\mathbb{R}),Q) \\
& f & \mapsto & \int f(x)\, N(\cdot,\sigma^2)(\mathrm{d}x).
\end{array}
\end{equation*}
I am interested in the eigenspace $E_1 := \mathrm{kernel(I-R)},$ in particular in the dimension of $E_1.$
Obviously, the indicator function $\mathbb{1}_{\mathbb{R}}: x \mapsto 1$ and the identity $\mathrm{id}_{\mathbb{R}}: x \mapsto x$ are both eigenfunctions to the eigenvalue $1,$ that is, $\mathbb{1}_{\mathbb{R}},\ \mathrm{id}_{\mathbb{R}} \in E_1.$ 
Are there more linearly independent eigenfunctions?
 A: If I understand correctly, your operator $R$ is the convolution operator with the Gauss–Weierstrass kernel. This is a Fourier multiplier with symbol $\lambda(\xi) = \exp(-\tfrac{1}{2} \sigma^2 |\xi|^2)$:
$$ \widehat{R f}(\xi) = \lambda(\xi) \hat f(\xi). $$
If $f$ is a tempered distribution, then $R f = f$ if and only if $$(\exp(-\tfrac{1}{2} \sigma^2 |\xi|^2) - 1) \hat{f}(\xi) = 0 ,$$ and this is equivalent to $\hat{f}$ being supported in $\{0\}$. This, in turn, implies that $f$ is a polynomial. By inspection, in dimension $1$ the eigenspace is indeed spanned by $f(x) = 1$ and $f(x) = x$. In higher dimensions, however, any harmonic polynomial (i.e. a polynomial $f$ such that $\Delta f = 0$) will do.
If one goes beyond tempered distributions, there are more solutions, even in dimension one. For example, $f(x) = e^{z x}$ is an eigenfunction corresponding to eigenvalue $$\lambda(z) = \exp(-\tfrac{1}{2} \sigma^2 z^2),$$ where $z$ is an arbitrary complex number. Choosing, for example, $$z = \sqrt{2 \pi} \sigma^{-1} (1 + i),$$ we get $\lambda(z) = \exp(-2 \pi i) = 1$, as desired.
If one insists on real-valued solutions, then $f(x) = \Re e^{z x} = e^{x \Re z} \cos(\Im z)$ works (as long as $\lambda(z)$ is real). Thus, to give a specific real-valued example,
$$ f(x) = e^{\sqrt{2 \pi} x / \sigma} \cos(\sqrt{2 \pi} x / \sigma) $$
is another eigenfunction with eigenvalue $1$.
