If $\lambda$ is a characteristic root of a non-singular matrix $A$, then prove that $det(A)/\lambda$ is a characteristic root of $Adj A$. Let $A$ be a $n\times n$ matrix. Then the characteristic equation is $\mathrm{det}(A-\lambda×I)=0$, with possibly characteristic roots of $A$ $\lambda_1,\dots,\lambda_n$. Let $A$ be regular, i.e. $\mathrm{det}(A)\neq0$. Now $\mathrm{Adj}(A)=\mathrm{det}(A)\mathrm{Inv}(A)$. I don't know how to find its characteristic roots.
Let $A$ be invertible. As you rightly remarked, then $\mathrm{adj}(A)=\mathrm{det}(A)A^{-1}$. We can even prove the following stronger proposition: 

Proposition: For $A$ invertible, $\lambda\neq 0$ is an eigenvalue of $A$ iff $\mathrm{det}(A)\lambda^{-1}$ is an eigenvalue of $\mathrm{adj}(A)$:

Proof:
$$
Av=\lambda v\Leftrightarrow \mathrm{adj}(A)(Av)=\mathrm{adj}(A)(\lambda v)\Leftrightarrow \mathrm{det}(A)A^{-1}(Av)=\mathrm{det}(A)A^{-1}(\lambda v)\Leftrightarrow \mathrm{det}(A)v=\mathrm{det}(A)\lambda A^{-1}(v)\Leftrightarrow\mathrm{det}(A)\lambda^{-1}v=\mathrm{det}(A)A^{-1}(v)\Leftrightarrow\mathrm{det}(A)\lambda^{-1}v=\mathrm{adj}(A)v
$$
$\Box$

You may rephrase this result in terms of roots of the characteristic polynomial, as you naturally have for $\lambda\neq 0$ that $\mathrm{det}(A-\lambda E_n)=0$ iff $\lambda$ is an eigenvalue of $A$ iff $\mathrm{det}(A)\lambda^{-1}$ is an eigenvalue of $\mathrm{adj}(A)$ iff $\mathrm{det}(\mathrm{adj}(A)-\mathrm{det}(A)\lambda^{-1}E_n)=0$.

Obviously, the condition $\lambda\neq 0$ is w.l.o.g. as $0$ is a eigenvalue of $A$ iff $\det(A-0E_n)=\det(A)=0$ iff $A$ is singular.
AX=aX where a is the eigen value of non singular matrix A 
(adj A)AX=adj A (aX)
|A|A^- (AX)= a (adj A)X
|A|X=a(adj A) X
|A|X/a=(adj A)X
(adj A)X=|A|X/a
=》|A|/a is the characteristic root of matrix adjA