Consider a pair of independent random variables $(V,Z)$ where $Z$ is standard normal. Now Supposesuppose that the following equality holds: for a given $a>0$ \begin{align} \mathbb{P}[ a V\le Z| V+Z]=\mathbb{P}[aV \ge Z| V+Z] \text{ a.s.} \end{align}
Question: Can we show that the above is true if and only if $V=\frac{1}{\sqrt{a}}Z'$ where $Z'$ is standard normal independent of $Z$?
To show the forward direction note that if we choose $V=\frac{1}{\sqrt{a}}Z'$, we get \begin{align} \mathbb{P} \left[ \sqrt{a} Z'- Z \le 0| \frac{1}{\sqrt{a}}Z'+Z \right]=\mathbb{P} \left[\sqrt{a} Z'- Z \ge 0| \frac{1}{\sqrt{a}}Z'+Z \right] \text{ a.s.} \end{align}
Now note that $ \sqrt{a} Z'- Z$ and $\frac{1}{\sqrt{a}}Z'+Z$ are independent, to see this observe that \begin{align} \mathbb{E} \left[(\sqrt{a} Z'- Z)(\frac{1}{\sqrt{a}}Z'+Z)\right]=0 \end{align}\begin{align} \mathbb{E} \left[(\sqrt{a} Z'- Z)(\frac{1}{\sqrt{a}}Z'+Z)\right]=0, \end{align} and recall that uncorolated is equivalent to independence for gaussian random variables.
Consequently, we have that \begin{align} \mathbb{P}[ \sqrt{a} Z'- Z \le 0]=\mathbb{P}[\sqrt{a} Z'- Z \ge 0] \end{align} which is true by summary.
The other direction appears to be hard or not true. I am not sure.
I was thinking of re-writing the equality condition as
\begin{align}
\mathbb{P} \left[ \left[\begin{array}{c} a\\-1 \end{array} \right] (V,Z) \le 0 | \left[\begin{array}{c} 1\\1 \end{array} \right] (V,Z) \right]=\mathbb{P} \left[ \left[\begin{array}{c} a\\-1 \end{array} \right] (V,Z) \ge 0 | \left[\begin{array}{c} 1\\1 \end{array} \right] (V,Z) \right],
\end{align}
and then either using some kind projection theorem or change of variable.