Let
\begin{align}
W=U+Z
\end{align}
where $Z$ is a standard normal and independent of random variable $U$. Suppose that for some given $a \in (0,1)$
\begin{align}
F_{U|W}(aw|w)=\frac{1}{2}, \forall w\in \mathbb{R}.
\end{align}
where $F_{U|W}(u|w)$ is the conditional cdf of $U$ given $W$, that is
\begin{align}
F_{U|W}(u|w)=P( U \le u| W=w)
\end{align}
Question: Show that the above holds if and only if $U$ is Gaussian. (Assume $U$ is zero mean)
Forward direction: One of the directions is easy if $U$ is Gaussian, since the sum is Guassian we have that \begin{align} F_{U|W}(t|w)=\Phi \left( \frac{t- E[U|W=w]}{\sqrt{Var(U|W=w)}}\right)=\Phi \left( \frac{t- b w}{\sqrt{b}} \right) \end{align} where $\Phi(\cdot)$ is the cdf of standard normal and we have used that $E[U|W=w]=bw$ and $\frac{E[U^2]}{E[U^2]+1}=Var(U|W=w)=b$.
Therefore, the condition reduces to \begin{align} \Phi \left( \frac{aw- b w}{\sqrt{b}} \right)=\frac{1}{2}, \forall w\in \mathbb{R}. \end{align} which is equivalent to \begin{align} aw- b w =0, \forall w\in \mathbb{R}, \end{align}
Therefore, if we choose $U$ to be Gaussian such that $a=b=\frac{E[U^2]}{E[U^2]+1}$, it will match the desired equation.
The question now is how to show that Gaussian is the only one that satisfies this.