Let $A$ and $B$ be $2n$-dimensional complex vector bundles and $\det A=\Lambda^{2n}(A)$ and $\det B=\Lambda^{2n}(B)$.
Can you prove $A\cong B $ if and only if $\det A\cong \det B $? Is it a correct proposition?
Note: Here $\cong$ is a bundle isomorphism.
For a simple way to get counterexamples, note that if $A\cong L_1\oplus\dots\oplus L_n$ splits a direct sum of line bundles, then $\det A\cong L_1\otimes \dots\otimes L_n$. So, for instance, let $X$ be any space with a class $\alpha\in H^2(X,\mathbb{Z})$ such that $\alpha^2\neq 0$ and let $L$ be the line bundle on $X$ with $c_1(L)=\alpha$. Letting $A=L\oplus L^{-1}$ and $B$ be the trivial rank $2$ bundle, then $\det A\cong L\otimes L^{-1}$ and $\det B$ are both trivial, but $A\not\cong B$ since $c_2(A)=c_1(L)c_1(L^{-1})=-\alpha^2\neq 0$ so $A$ is nontrivial.