Let $C$ be a complete $\infty$-category.

Let $U:Fun(BC_n,C)\to C$ denote the forgetful functor, $\mathrm{CoInd}$ its right adjoint, and $(-)^{triv}$ the functor given by precomposition along $BC_n\to *$.

Then we have a canonical equivalence $U(X^{triv})\to X$ which yields, by adjunction, a map $X^{triv}\to \mathrm{CoInd}(X)$ which is $C_n$-equivariant.

Apply this to $C= Cat_\infty$ and $X=Sp$ yields a $C_n$-equivariant map $Sp \to \mathrm{CoInd}(Sp)$. Now $\mathrm{CoInd}(Sp) = Sp^{\times n}$ with the permutation action.

Now $Sp$ can be canonically seen as a commutative monoid in $Cat_\infty$, that is, a certain type of functor $Fin_*\to Cat_\infty$, which we can then obviously restrict to $Fin$ to get $Fin\to Cat_\infty$, informally given by $n\mapsto Sp^{\times n}$.

In particular, we get a $\Sigma_n$-equivariant map $Sp^{\times n}\to Sp$ corresponding to the smash product, and the action of $\Sigma_n$ on $Sp^{\times n}$ restricts to the permutation action of $C_n$

You can prove this by dealing with the universal case.

Another way to do that, which certainly agrees, is to note that in $CAlg(Cat_\infty)$, products and coproducts agree, i.e. it is preadditive, and so $\mathrm{CoInd}(Sp)\simeq \mathrm{Ind}(Sp)$, in particular you get for free a $C_n$-equivariant map $\mathrm{CoInd}(Sp)\simeq \mathrm{Ind}(Sp)\to Sp$ (from $Sp\to U(Sp^{triv})$) which is also given by smash product.

Anyways, it follows that both $Sp\to Sp^{\times n}$ and $Sp^{\times n}\to Sp$ are $C_n$-equivariant

Your left Kan extension construction will not work. Left Kan extending along $*\to BG$ is left adjoint to the forgetful functor, i.e. it is induction - when composed with the forgetful map, this looks like $\bigoplus_{g\in G}$, so if you left Kan extend $X\mapsto X\otimes ... \otimes X$, you will get $\bigoplus_{g\in C_p}X\otimes... \otimes X$, and no permutation action.

As Harry already pointed out, the answer to your side question is "yes", the inclusion has both a left and a right adjoint, in particular it preserves limits and colimits.

Let $C$ be a complete $\infty$-category.

Let $U:Fun(BC_n,C)\to C$ denote the forgetful functor, $\mathrm{CoInd}$ its right adjoint, and $(-)^{triv}$ the functor given by precomposition along $BC_n\to *$.

Then we have a canonical equivalence $U(X^{triv})\to X$ which yields, by adjunction, a map $X^{triv}\to \mathrm{CoInd}(X)$ which is $C_n$-equivariant.

Apply this to $C= Cat_\infty$ and $X=Sp$ yields a $C_n$-equivariant map $Sp \to \mathrm{CoInd}(Sp)$. Now $\mathrm{CoInd}(Sp) = Sp^{\times n}$ with the permutation action.

Now $Sp$ can be canonically seen as a commutative monoid in $Cat_\infty$, that is, a certain type of functor $Fin_*\to Cat_\infty$, which we can then obviously restrict to $Fin$ to get $Fin\to Cat_\infty$, informally given by $n\mapsto Sp^{\times n}$.

In particular, we get a $\Sigma_n$-equivariant map $Sp^{\times n}\to Sp$ corresponding to the smash product, and the action of $\Sigma_n$ on $Sp^{\times n}$ restricts to the permutation action of $C_n$

You can prove this by dealing with the universal case.

Another way to do that, which certainly agrees, is to note that in $CAlg(Cat_\infty)$, products and coproducts agree, i.e. it is preadditive and so induction and co-induction agree. In particular you get for free a $C_n$-equivariant map $\mathrm{CoInd}(Sp)\to Sp$ (from $Sp\to U(Sp^{triv})$) which is also given by smash product.

Anyways, it follows that both $Sp\to Sp^{\times n}$ and $Sp^{\times n}\to Sp$ are $C_n$-equivariant

Your left Kan extension construction will not work. Left Kan extending along $*\to BG$ is left adjoint to the forgetful functor, i.e. it is induction - when composed with the forgetful map, this looks like $\bigoplus_{g\in G}$, so if you left Kan extend $X\mapsto X\otimes ... \otimes X$, you will get $\bigoplus_{g\in C_p}X\otimes... \otimes X$, and no permutation action.

As Harry already pointed out, the answer to your side question is "yes", the inclusion has both a left and a right adjoint, in particular it preserves limits and colimits.