If $a$ is a quadratic residue of odd prime $p$, then is it the case that $a^{-1}$ is also a quadratic residue? If $a$ is a quadratic residue of odd prime $p$, then is it the case that $a^{-1}$ is also a quadratic residue ?

I notice quadratic residues of odd prime $13$ : 
$$1^2\equiv 12^2\equiv 1\\2^2\equiv 11^2\equiv 4\\3^2\equiv 10^2\equiv 9\\4^2\equiv 9^2\equiv 3\\5^2\equiv 8^2\equiv 12\\6^2\equiv 7^2\equiv 10$$
$4^{-1}=10\\9^{-1}=3\\1^{-1}=1\\12^{-1}=12$ 
Is this true always ? How to go about proving ? thanks! 

Maps like $\,x\mapsto x^{-1}$ that are multiplicative $\,h(ab) = h(a)h(b)$ preserve squares $\,h(a^2)= (h(a))^2$
