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.