If A changes to $4A$, what is the change in SVD?
$$A^tA=(U \Sigma V)^t(U \Sigma V)$$
So, if we change $A$ to 4A,
$$(4A^t)(4A)=16(A^tA)=16(U \Sigma V)^t(U \Sigma V)$$
It seems like that SVD decomposition is multiplied by 16, but I am not sure...
Alsom
What is the SVD for $A^T$ and $A^{-1}$?
When $A=U \Sigma V^T$, $4A=U (4 \Sigma) V^T$. Now $U$ and $V$ are still unitary and $4 \Sigma$ is still diagonal with nonnegative entries, so this is indeed the SVD of $4A$. Note that this would not work as written with $-A$, since $-\Sigma$ doesn't have nonnegative entries.
You seem to have confused the SVD of $A$ with the eigendecompositions of $A^T A$ and $A A^T$. These are related (in the sense that you can easily compute one from the other) but not the same.