Does $\mathrm{adj}(A)=\mathrm{adj}(B)$ imply $A=B$?

No.

The adjugate matrix is basically the wedge power $\wedge^{n-1} A$ if $n$ is the dimension of the underlying vector space. If the underlying field has nontrivial $(n-1)$-th roots of unity, then we can multiply $A$ by one of them without changing the adjugate matrix. Also, the adjugate matrix of any matrix of rank $< n-1$ will be $0$.

Not necessarily. It is given: $adj(A)=A^{-1}|A|=adj(B)=B^{-1}|B|$. Multiply both sides by $A$, then $B$ to get: $A=\frac{|A|}{|B|}B$. It implies: $$|A|=\left(\frac{|A|}{|B|}\right)^n|B| \Rightarrow \left(\frac{|A|}{|B|}\right)^{n-1}=1\Rightarrow \begin{cases} |A|=|B|, \ \ \ \ \ if \ n \ is \ even \\ |A|=\pm|B|, \ \ if \ n \ is \ odd \end{cases}.$$

Indeed

if $A^{-1}=B^{-1} $ then $AA^{-1}B=AB^{-1}B$ i.e. $B=A $.

If $A,B$ are non-invertible then $\det(A)=\det(B)=0$ and $\text{adj}(A)A=\det(A)I=0$ and $\text{adj}(B)B=\det(B)I=0$.

This can be satisfied without $A=B$.

Hence counterexamples for supposed theorem are, for example

$\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \ \end{bmatrix}$, $\begin{bmatrix} 0 & -1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \ \end{bmatrix}$.

Here of course $\text{adj}(A)=\text{adj}(B)$.