Doesn't this follow from dependence on initial data?
Consider the flow mapping $\Phi(t,X)$ which solves
$$ \frac{d}{dt}\Phi(t,X) = - \nabla V(\Phi(t,X)) $$
so taking the derivative in $X$ we have
$$ \frac{d}{dt} \partial_X \Phi(t,X) = - \nabla^2 V(\Phi(t,X)) \cdot \partial_X \Phi(t,X) \\= - \nabla^2 V(X) \cdot \partial_X \Phi(t,X) + O(t) \cdot \partial_X \Phi(t,X)$$
So if $-\nabla^2 V(X_0)$ has negative eigenvalue $-\lambda_0$ with eigenvector $v_0$, taking the partial in the $v_0$ direction gives
$$ \partial_{v_0} \Phi(t,X_0) = e^{\lambda_0 t} v_0 + O(t^2) $$
For $t>0$ sufficiently small you guarantee that $$ |\partial_{v_0} \Phi(t,X_0) | \geq (1 + \frac{\lambda_0}{2}t) |v_0| $$ showing that the solution map cannot be 1 Lipschitz.