A spectrum $X$ is dualizable if the natural map $$Map(X,\mathbb S) \wedge X \rightarrow Map(X,X)$$ is an equivalence of spectra. This is equivalent to having evaluation and coevaluation maps in the stable homotopy category $$ X \wedge DX \rightarrow \mathbb S $$ $$ \mathbb S \rightarrow DX \wedge X $$ for which the usual composites $$ X \rightarrow X \wedge DX \wedge X \rightarrow X $$ $$ DX \rightarrow DX \wedge X \wedge DX \rightarrow DX $$ give the identity in the homotopy category (cf. Lewis-May-Steinberger III.1.2). It is also well-known that this implies that $X \rightarrow D(DX)$ is an equivalence of spectra (LMS III.1.3(i)), but my question is
Is the converse true? Does $X \overset\sim\rightarrow D(DX)$ imply that $X$ is dualizable?
I understand that everything I have said holds in an arbitrary closed symmetric monoidal category, but I am willing to consider arguments that only work for spectra (or $R$-module spectra).