Edit: Here's a simpler method for producing counterexamples generalizing the argument below, we end up only needing to discuss the Picard group. If $P$ is a f.g. projective module of rank $n$ in the sense that $\Lambda^n(P)$ is invertible, then the same is true of $I \otimes P^{\ast}$ where $I$ is invertible, and
$$\Lambda^n(I \otimes P^{\ast}) \cong I^{\otimes n} \otimes \Lambda^n(P)^{-1}.$$
So to construct a f.g. projective module $P$ of rank $n \ge 3$ which is not isomorphic to $I \otimes P^{\ast}$ it suffices to find $P$ such that the class of $\Lambda^n(P)^{\otimes 2}$ in the Picard group is not divisible by $n$. If $P \cong I_1 \oplus \dots \oplus I_n$ where each $I_j$ is invertible then $\Lambda^n(P) \cong I_1 \otimes \dots \otimes I_n$ so the class of $\Lambda^n(P)$ can be arbitrary, so to write down, say, a Noetherian counterexample it suffices to find a number field $K$ such that the ideal class group of $\mathcal{O}_K$ is, say, cyclic of order $3$, then take $n = 3$ and $\Lambda^n(P)$ to be nontrivial. Explicitly we can take $K = \mathbb{Q}(\sqrt{-23})$.
This approach doesn't answer your question about what happens if the Picard group is trivial. I think the characteristic class argument below ought to generalize (although we would need to look at $c_3$ or higher rather than $c_1$) but I'm not sure.
No. Using the Serre-Swan theorem we can find counterexamples using vector bundles, where we can distinguish a bundle $V$ from its dual $V^{\ast}$ using characteristic classes. For example, the first Chern class of the tangent bundle $T$ of $\mathbb{CP}^3$ is
$$c_1(T) = 4 \alpha$$
where $\alpha \in H^2(\mathbb{CP}^3) \cong \mathbb{Z}$ is a generator of the cohomology ring. This means the first Chern class of the dual is
$$c_1(T^{\ast}) = - 4 \alpha$$
(which can be proven using the splitting principle). If $L$ is a line bundle with Chern class $c_1(L) = k \alpha \in H^2(\mathbb{CP}^3)$ then the first Chern class of $L \otimes T^{\ast}$ is
$$c_1(L \otimes T^{\ast}) = (3k - 4) \alpha$$
(this is also a calculation using the splitting principle), and since $4 \not \equiv -4 \bmod 3$ there is no value of $k$ that makes the first Chern classes match up and hence no $L$ such that $T \cong L \otimes T^{\ast}$. To be explicit, the ring we're considering f.g. projective modules over here is the ring $C^{\infty}(\mathbb{CP}^3, \mathbb{C})$ of complex-valued smooth functions on $\mathbb{CP}^3$ (we could also consider continuous functions).
One reason to see why we had to go up to rank $3$ to get a counterexample is that the statement is actually true for a f.g. projective module $P$ of rank either $1$ or $2$, where by "rank $n$" I mean that $\Lambda^n(P)$ is invertible. If $P$ is itself invertible then we can take $I = P^{\otimes 2}$, whereas if $\Lambda^2(P)$ is invertible then the wedge product is a nondegenerate bilinear form
$$P \otimes P \to \Lambda^2(P)$$
so we can take $I = \Lambda^2(P)$. So if a counterexample exists and has constant rank its rank has to be $\ge 3$.
