Does every invertible matrix A has a matrix B such that A=Adj(B)? I'm trying to understand if it's always true, always true over $\mathbb C$ or never true.
I know that if $A$ is invertible, than there exists $A^{-1}$.
$$A=\frac{1}{det (A^{-1})}Adj(A^{-1})$$
So I have an adjoint matrix multiplied by a scalar, but how do I know if the result is an adjoint by itself?

The case $n=1$ is immediate. For the case $n\ge 2$, by user1551's comment,
$${\rm adj}({\rm adj}{(A)})=(\det A)^{n-2}A.$$
Also, there is a fact that ${\rm adj}(kA)=k^{n-1}{\rm adj}(A)$ for any scalar $k\in\mathbb{C}$. Thus we have
\begin{align}
{\rm adj}((\det A)^{-\frac{n-2}{n-1}}{\rm adj}{(A)})
&=(\det A)^{-(n-2)}{\rm adj}({\rm adj}{(A)})\\
&=(\det A)^{-(n-2)}(\det A)^{n-2}A\\
&=A,
\end{align}
and the result follows by taking $B=(\det A)^{-\frac{n-2}{n-1}}{\rm adj}{(A)}$.
