The question is for a given variance $\sigma_0^2$, to find a probability distribution $p(x)$, whose variance is $\sigma_0^2$, that maximizes the overlap between the distribution and its Gaussian convolution of variance $\sigma_1^2$, i.e., \begin{align} \int_{-\infty}^\infty\, dx\ p(x) \int_{-\infty}^\infty\, dx'\ h(x') p(x-x'), \end{align} where $h(x)=\dfrac{e^{-x^2/(2\sigma_1^2)}}{\sqrt{2\pi\sigma_1^2}}$.
My guess is a Gaussian distribution with a given variance, but I don't know how to prove.