First, having seen the edited version of Martin's question, let's quickly dispose of the construction of the free symmetric monoidal category generated by a category $C$. Objects are tuples $(x_1, \ldots, x_n)$ of objects of $C$. Morphisms are labeled permutations, where permutations are conveniently visualized as string diagrams, each string being labeled by a morphism in $C$. To compose such labeled diagrams, just compose the string diagrams, composing the labels of strings in $C$ along the way.
The free symmetric monoidal category $Sym(M)$ on a monoidal category $M$ is formed from the free symmetric monoidal category $S (U M)$ on the underlying category $U M$ by adjoining isomorphisms $\phi_{x_1, \ldots, x_n}: (x_1, \ldots, x_n) \to x_1 \otimes \ldots \otimes x_n$, where the $x_i$ are objects of $M$, the expression $(x_1, \ldots, x_n)$ is the formal monoidal product in $S(UM)$, and $x_1 \otimes \ldots \otimes x_n$ is the tensor product in $M$.
More precisely, define the objects of $Sym(M)$ to be tuples $(x_1, \ldots, x_n)$ of objects of $M$. Define morphisms $(x_1, \ldots, x_n) \to (y_1, \ldots, y_m)$ to be equivalence classes of pairs $(p, f)$ where $p$ is a permutation on $n$ elements, and $f$ is an $M$-labeled planar forest of $m$ rooted trees. (You should think here of the free multicategory generated by a category.) Formally, a planar forest can be described as a functor $[n]^{op} \to \Delta$ where $n$ is the category $1 \leq \ldots \leq n$ and $\Delta$ is the category of finite (possibly empty) ordinals. Under the obvious way of drawing such forests, the ordered list of leaves of the forest is labeled by $(x_{p(1)}, \ldots, x_{p(n)})$, and the ordered list of roots by $(y_1, \ldots, y_m)$. Edges are labeled by objects of $M$, and each internal node with $k$ inputs labeled (in order) by $m_1, \ldots, m_k$ and output $m$ is labeled by a morphism $f: m_1 \otimes \ldots \otimes m_k \to m$.
There is an evident way, using the monoidal structure of $M$, of evaluating such a labeled forest $f$ as a morphism $ev(f): x_{p(1)} \otimes \ldots \otimes x_{p(n)} \to y_1 \otimes \ldots \otimes y_m$ in $M$. We consider two arrows $(p, f)$ and $(p', f')$ to be equivalent if $p = p'$ as permutations and $ev(f) = ev(f')$.
Now we define composition of pairs. The main idea is to rewrite the composition of a forest followed by a permutation,
$$(x_1, \ldots, x_n) \stackrel{(1, f)}{\to} (y_1, \ldots, y_m) \stackrel{(q, 1)}{\to} (y_{q(1)}, \ldots, y_{q(m)}),$$
into a form $(p, f')$ forced by the naturality requirement of the symmetry isomorphism. Namely, if $\bar{x}_i$ is the tuple of leaves of the tree whose root is $y_i$, then we have a block permutation $bl(q)$ taking $(\bar{x}_1, \ldots, \bar{x}_m)$ to $(\bar{x}_{q(1)}, \ldots, \bar{x}_{q(m)})$. The permutation $q$ can also be applied to the $m$ trees of the forest by reordering the trees, yielding a new forest $\mathrm{perm}_q(f)$, and the composition $(q, 1) \circ (1, f)$ is defined to be $(bl(q), \mathrm{perm}_q(f))$. Then, if we have $(p, f): (x_1, \ldots, x_n) \to (y_1, \ldots, y_m)$ and $(q, g): (y_1, \ldots, y_m) \to (z_1, \ldots, z_p)$, we define their composite to be
$$(bl(q) \circ p, g \circ \mathrm{perm}_q(f))$$
where the first component is by composing permutations and the second is by the usual way of composing forests by plugging in roots of one planar forest of trees into leaves of another.
Remaining details that all this works will be left to the reader. I will remark that the key rewriting procedure above is an instance of a kind of distributive law; there are some more details to this effect in some notes on my nLab web; see here.
$$Fun^{\otimes}(\mathbb Z/2, C) \simeq map_{E_1}(\mathbb Z/2, C^\simeq) \simeq map_{E_k}(\mathrm{Ind}_{E_1}^{E_k}\mathbb Z/2, C^\simeq)$$
where $\mathrm{Ind}_{E_1}^{E_k}$ denotes the left adjoint to the forgetful functor.
So $Inv$ is representable, and the natural $E_{k-1}$-structure (see my comments for why I wrote $E_{k-1}$ and not $E_k$ - it is possible that in this special case too we could get $E_k$, but I don't see a reason why, and what I wrote works for any $E_1$-space $X$) on this space gives $\mathrm{Ind}_{E_1}^{E_k}\mathbb Z/2$ a natural co-$E_{k-1}$-structure (in $E_k$-spaces - with the coproduct monoidal structure).
Now does the space $\mathrm{Ind}_{E_1}^{E_k}\mathbb Z/2$ have a reasonably concrete description ? I think it's something like a bar construction $Bar(E_1,E_k, \mathbb Z/2)$ so you can get an explicit description involving the space of little $k$-disks, but I'm not entirely sure you can get much better. I'd love to hear about a better description.
Best Answer
I assume that with ``monoidal ∞-groupoid'' you mean an $E_1$-space. In this case the answer is yes. It is well known that $E_1$-spaces can be modeled by functors
$$X:\Delta^{\mathrm{op}}\to \operatorname{Space}$$ satisfying the Segal conditions. Now if you are given an $\infty$-category $\mathcal{C}$ you can define a simplicial space $$s(\mathcal{C}): [n]\mapsto\operatorname{Map}_{\operatorname{Cat}_∞}(\Delta^n,\mathcal{C})\,.$$ In fact this functor is fully faithful and identifies $\operatorname{Cat}_∞$ with the category of complete Segal spaces. For the following we won't need all this though - we will use only that it takes values in Segal spaces (which follows immediately from $\Delta^n\amalg_{\Delta^0} \Delta^m\simeq\Delta^{n+m-1}$ in $\operatorname{Cat}_∞$).
Now let $x\in\mathcal{C}$ be an object of $\mathcal{C}$. Then we can define the simplicial space $$ \operatorname{End}_{\mathcal{C}}(x):\Delta^{\mathrm{op}}\to \operatorname{Space}\qquad [n]\mapsto \{x\}\times_{\operatorname{Map}(\{0,\dots,n\},\mathcal{C})} \operatorname{Map}_{\operatorname{Cat}_∞}(\Delta^n,\mathcal{C})\,.$$ That is it sends $[n]$ to the (∞-)groupoid of functors $F:\Delta^n\to \mathcal{C}$ that sends all objects to $x$. It is easy now to see that $\operatorname{End}_{\mathcal{C}}(x)$ satisfies the Segal conditions and so it is an $E_1$-space.
This takes care of your preliminary question. To go back to your main question, the functor $(\mathcal{C},x)\mapsto \operatorname{End}_{\mathcal{C}}(x)$ obviously preserves all limits and filtered colimits, and so it has a left adjoint $B$ exactly as you wanted. To get a more ``concrete'' description $B$ sends an $E_1$-space $X$ to the ∞-category corresponding to the completion of $X$ seen as a Segal space. That is $$BX:=\int^{[n]\in\Delta^{\mathrm{op}}} X([n])\times \Delta^n$$ where the coend is computed in $\operatorname{Cat}_∞$.
With more care one can show that $B:E_1-\operatorname{Space}\to(\operatorname{Cat}_∞)_{\Delta^0/}$ is fully faithful with essential images those arrows $\Delta^0\to\mathcal{C}$ that are essentially surjective (that is such that $\mathcal{C}$ has only one equivalence class of objects). Indeed this is a special case of the equivalence between the ∞-category of Segal spaces and the ∞-category of flagged ∞-categories.