If a dualizing object exists, there is a bijection between isomorphism classes of dualizing objects and isomorphism classes of $\otimes$-invertible objects (i.e. the Picard group), given by tensoring your favorite dualizing object by a $\otimes$-invertible object. So the groupoid of $\ast$-autonomous structures, if nonempty, is equivalent to the groupoid of $\otimes$-invertible objects (canonically as soon as one chooses a basepoint -- it's a torsor over the grouplike symmetric monoidal groupoid of $\otimes$-invertible objects).
One direction:
First let's check that if $D$ is dualizing and $L$ is $\otimes$-invertible (with inverse $L^\vee$), then $L\otimes D$ is dualizing. Write $[,]$ for the internal hom (sorry, I seem to be changing all of your notation :). Then
$\begin{align*}
[[A, L\otimes D], L\otimes D]
&= [L^\vee \otimes [A, L\otimes D], D] \\
&= [L^\vee \otimes L \otimes [A,D],D] \\
&= [[A,D],D] \\
&= A
\end{align*}$
I suppose I should verify that the above isomorphism is the canonical morphism $A \to [[A, L\otimes D], L\otimes D]$, but since all the isomorphisms used were canonical, maybe I'll just wave my hands and ask rhetorically, "what else could it be?".
The converse:
In fact, it's always the case that any two dualizing objects differ by tensoring by a $\otimes$-invertible object. Here's a proof.
First note that if $D$ is dualizing, then $[D,D] = I$. To see this, it suffices to check that $[[D,D],D] = [I,D]$ because $[-,D]$ is a contravariant equivalence of categories. But both sides are $D$, so this is the case.
Now if $D,D'$ are both dualizing, I claim that $[D,D']$ is $\otimes$-invertible, with inverse $[D',D]$. To see this, it suffices by symmetry to show that $[D,D'] \otimes [D',D] = I$. To check this, it suffices to check that $[[D,D'] \otimes [D',D],D] = [I,D]$. The lefthand side simplifies to $[[D,D'], [[D',D],D]] = [[D,D'],D'] = D$ where we have curried, and then used the dualizing property of both $D$ and $D'$. Of course, this is the same as the righthand side.
Finally, I claim that $D \otimes [D,D'] = D'$. To see this, it suffices to check that $[D \otimes [D,D'], D'] = [D',D']$. We've already seen that the righthand side is $I$ in the first bullet. And the lefthand side simplifies to $[D, [[D,D'],D']] = [D,D] = I$.
So $D$ and $D'$ differ by tensoring by the $\otimes$-invertible object $[D,D']$.
And of course, since tensoring with a $\otimes$-invertible object is an equivalence of categories, the action map $L \mapsto L \otimes D$ is fully faithful; we've just seen it's essentially surjective, so it's an equivalence of groupoids.
Endnote:
More than once I've found myself questioning the equation $[A,L\otimes D] = L \otimes [A,D]$ used in the second line of the forward direction so let me just record the proof here for my own benefit:
$\begin{align*}
Hom(X,[A,L\otimes D]) &= Hom(X \otimes A, L\otimes D) \\
&= Hom(X \otimes A \otimes L^\vee, D) \\
&= Hom(X \otimes L^\vee, [A,D]) \\
&= Hom(X, L \otimes [A,D])
\end{align*}$
and conclude by Yoneda. So this isomorphism holds for any dualizable $L$ and arbitrary $A,D$.
