Here's some background and notation:

Let $G/K$ be a symmetric space of non-compact type. For concreteness, assume $G$ is in fact a classical simple real Lie group such as SL,SO, or Sp, and $K$ is a maximal compact subgroup in $G$.

The tangent space to $p_{0}=1 \in G/K$ is identified with $\frak{p}$ where $\frak{p}$ is an orthogonal complement to the Lie algebra of $K$ inside the Lie algebra of $G$, with respect to the trace pairing on the latter. Furthermore, the exponential map $\exp: {\frak p} \rightarrow G$ followed by the quotient map $G \rightarrow G/K$ is a diffeomorphism. Put another way, given a point $p \in G/K$, there is a unique $x \in \frak{p}$ so that $\exp(x)=p$, and the unique length-parametrized geodesic connecting $p_{0}$ to $p$ is $\exp(t x/|x|)$.

I would like to understand the "equidistant hypersurface" $H_{p_{0},p}$ between $p_{0}$ and $p$, consisting of those points $q$ in $G/K$ that are equidistant from $p_{0}$ and $p$.

My question is whether this equidistant hypersurface $H_{p_{0},p}$ can be obtained by exponentiating the affine hyperplane in $\frak{p}$ that passes through the centre of and is orthogonal to the line connecting $0$ and $x$, where $\exp(x)=p$.

This seems intuitively correct, but sometimes intuition is misleading in higher rank symmetric spaces.

EDIT: The answer is no for the question as stated (see answers below), so I would like to revise the question to the following:

Explain why the bisector between $p_{0}$ and $p$ is obtained by exponentiating the hyperplane orthogonal to $[p_{0},p]$ in the tangent space at the midpoint $m \in [p_{0},p]$.

Explain whether the following is true: we can compute the bisector between $p_{0}$ and $p$ by looking at the whole geodesic line through $p_{0}$ and $p$, computing the bisector between $p$ and $-p$ by exponentiating from the tangent space at $p_{0}$, and then translating this bisector along the geodesic from $p_{0}$ to the midpoint of $[p_{0},p]$.