Let $\mathcal{V}$ be a symmetric monoidal $\infty$.category, $A$ be an algebra in $\mathcal{V}$ and let $\mathcal{M}$ be an $\infty$-category with a left action of $\mathcal{V}$. It is possible define the left $\infty$-category of left $A$-module objects of $\mathcal{C}$ $LMod_{A}(\mathcal{C})$ (for a detailed definition see Definition 4.2.1.13 in Higher Algebra:https://people.math.harvard.edu/~lurie/papers/HA.pdf).
My question is: what does the mapping space of $LMod_{A}(C)$ look like?
My guess is that it is a colimit (probably it is a Bar- Costruction) of the following diagram: Let $(a,m_{a}:A\otimes a\to a),(b,m_{b}:A\otimes b\to b)\in LMod_{A}(\mathcal{C}) $
$Map_{LMod_{A}(\mathcal{C})}(a,b)\cong colimit( Map_{\mathcal{C}}(a,b)\rightrightarrows Map_{\mathcal{C}}(A\otimes a,b))\rightrightarrows\to Map_{\mathcal{C}}(A\otimes A\otimes a,b)\dots ) $ where with the token $\rightrightarrows \to$ I denote three parallel arrows.
The parallel two arrows $\rightrightarrows$ are: one precomposes with $m_{a}$ and the second one sends an arrows $a\to b$ in $A\otimes a\xrightarrow{A\otimes f} A\otimes b\xrightarrow{m_{b}} b$. And, as the number of arrows increases, the arrows will be defined similarly
In case, I would also like a reference.
I think you meant to use a limit instead of a colimit, i.e. to propose $$ \mathrm{map}_{\mathrm{LMod}_A(\mathcal{C}}(a,b)\simeq \mathrm{lim}(\mathrm{map}_{\mathcal{C}}(a,b)\rightrightarrows \mathrm{map}_{\mathcal{C}}(A\otimes a,b))\rightrightarrows\to \mathrm{map}_{\mathcal{C}}(A\otimes A\otimes a,b)\dots ) $$ (If you think about your intuition for limits and colimits, this makes more sense, and if you check the formula in the 1-category of modules over usual rings, then you also have this limit and not a colimit.)
This formula holds true. I will sketch the argument. The main input is Proposition 4.2.2.12 in Higher Algebra, which gives you a fully faithful functor from $\mathrm{LMod}(\mathcal{C})$ to a certain subcategory of the functor category $\mathrm{Fun}(\Delta^\mathrm{op}\times[1], \mathcal{C}^\otimes)$. Here, $[1]$ denotes the one-arrow category $0\to 1$. After fixing the algebra $A$, we are interested in natural transformations between functors $\Delta^\mathrm{op}\times [1] \to\mathcal{C}^\otimes$ that satisfy the properties listed in Definition 4.2.2.10, and for which the restriction $\Delta^\mathrm{op}\times\{1\}\to\Delta^\mathrm{op}\times[1]\to\mathcal{C}^\otimes$ encodes $A$.
Now, we use that mapping spaces in functor categories are computed via a certain end. Namely, given two functors $F,G\colon\mathcal{A}\to\mathcal{B}$ of $\infty$-categories, there is an equivalence $\mathrm{Nat}(F,G)\simeq\int_{a\in A}\mathrm{map}_\mathcal{B}(Fa,Ga)$. It is now a matter of figuring out how much of the natural transformation between the functors $\Delta^\mathrm{op}\times [1] \to\mathcal{C}^\otimes$ specifying your two $A$-modules is already fixed (namely, restricting to $\Delta^\mathrm{op}\times\{1\}$, the natural transformation has to act as the identity), and how many diagrams will automatically commute anyway (namely the diagrams in which we ask for our module homomorphism $f\colon a\to b$ to respect the multiplications that happen internally to $A$; the only diagrams that do not automatically commute are those in which we ask for $f$ to respect the action of $A$ on the modules). Moreover, we see that the maps $A\otimes\ldots\otimes A\otimes a\to A\otimes\ldots\otimes A\otimes b$ have to be of the form $A\otimes\ldots\otimes A\otimes f$. All this implies that the end formula that computes the space of natural transformations will in our case compute the same thing as the limit formula above.