Suppose you put a strictly positive (supported on all $\mathbb{R}^n$) probability measure $\psi$ on $\mathbb{R}^n$. Suppose its density has only one local maximum from which the density decreases in every direction. Let $K$ be a convex body with mean $0$, i.e.
$$\int_K x\ d\psi(x)=0$$
Let $0\neq u\in\mathbb{R}^n$. I want proof that the shifted convex body $K+u$ does not have mean $0$. Seemingly, it would usually be the case that $$u\cdot \int_{K+u}x\ d\psi(x) > 0$$
Of course, if $\psi$ is uniform on $K\cup K+u$, then the claim is true and the mean would shift by exactly $u$.
However, to show the necessity of the local maximum condition, consider if $K$ is a square based cone in $\mathbb{R}^3$
$$\{-1\le z\le 3-\max(|x|,|y|)\}$$
with $\psi=\psi^K+\psi^{K^c}$ (breaking up supports) uniform inside $K$ and $u=(0,0,0.5)$, we may construct $\psi^{K^c}$ for a counterexample. Pick the conditional distributions $\psi^{K^c}_{z=a}$ to be symmetric with weight $f(z)$ (to be determined), then in order for the mean to remain at $0$, we need
$$-\int_{-1}^{-.5}z(3-z)^2dz=\int_{-.5}^3z[(3.5-z)^2-(3-z)^2]f(z)dz+\int_3^{3.5}z(3.5-z)^2f(z)dz$$
You can weight $f(z)$ heavily enough between $z=-0.5$ and $z=0$ to achieve this. It does look like it would be an unhappy distribution, so there is hope for the proposition as stated.
If I am reading the question correctly I don't think what you asking to prove is true. Suppose the measure $\Psi$ is concentrated at $0$, or, more generally, has support that lies entirely inside the intersection of $K$ and $K+u $.
With the newly added hypothesis that the support is all of $\mathbb{R}^n$ I think the result is true. It's clear in dimension $1$. That implies that the mean must move in the direction of the offset $u$.