Let $k$ be a field and let $\mathcal{C}=\mathbf{StLin}_k$ be the $\infty$-category of stable infinity categories enriched over the $\infty$-category $\mathbf{Vect}_k$, regarded as a symmetric monoidal $\infty$-category with unit object $\mathbf{Vect}_k$.

The forgetful functor $\operatorname{CAlg}(\mathcal{C}) \to \mathcal{C}$ admits a left adjoint $Sym^*: \mathcal{C} \to \operatorname{CAlg}(\mathcal{C})$. Now let $\mathcal{C}[z]=Sym^*(\mathbf{Vect}_k)$ - i.e. $\mathcal{C}[z]$ is the free stable symmetric monoidal category generated by $\mathbf{Vect}_k$.

My question is: Is the $\infty$-category $\mathcal{C}[z]$ compactly generated in $\mathcal{C}$? (Recall that a category $\mathcal{A}$ is compactly generated if there exists a subcategory $\mathcal{A}_0 \subseteq \mathcal{A}$ and an equivalence $\mathcal{A} \simeq Ind(\mathcal{A}_0)$.)

More generally: If $\mathcal{D} \in \mathcal{C}$ is compactly generated, is $Sym^*(\mathcal{D})$ also compactly generated?

My idea was to show that the map $\mathbf{Vect}_k \to \mathcal{C}[z]$ corresponding to the identity:$ \mathcal{C}[z] \to \mathcal{C}[z]$ under the adjunction above identifies the image of $k$ with a compact generator of $\mathcal{C}[z]$, but I'm not sure if this would work.