If the Tate construction vanishes for all trivial $G$-actions, then does it vanish for all $G$-actions? Let $\mathcal{C}$ be a semiadditive $\infty$-category, complete and cocomplete, and let $G$ be a finite group. Then for any $X \in Fun(BG,\mathcal{C})$, there is a norm map $N_X: X_G \to X^G$. For each $X \in \mathcal{C}$, we have the constant $X^{triv} \in Fun(BG,\mathcal{C})$. Suppose that $N_{X^{triv}}$ is an equivalence for each $X \in \mathcal C$. Then does it follow that $N_X$ is an equivalence for each $X \in Fun(BG,\mathcal{C})$?
Evidence that this might be true comes from thinking in terms of ambidexterity: the question is whether n-ambidexterity can be checked on trivial objects when $n=1$. Note that this is the case when $n=(-2),(-1)$ (vacuously), or $n=0$ (If a pointed category $\mathcal C$ has finite coproducts and finite products, and if $X \vee X \to X \times X$ is an equivalence for each $X \in \mathcal C$, then $X \vee Y \to X \times Y$ is an equivalence for each $X,Y \in \mathcal C$).
Really, I'm interested in the question for all $n \in \mathbb N$:
Question: Let $\mathcal C$ be an $(n-1)$-ambidextrous $\infty$-category, complete and cocomplete. Let $B$ be an $n$-truncated space with finite homotopy groups. Then for all $X \in Fun(B,\mathcal C)$, there is a norm map $N_X: \varinjlim X \to \varprojlim X$. Suppose that this map is an equivalence when $X$ is constant. Then is $N_X$ an equivalence for all $X$?
Perhaps this doesn't hold "locally" (i.e. for a fixed $B$). Then can we at least say that if $N_X$ is an equivalence for all constant $X$ and all $n$-trucated, $\pi$-finite $B$, then $\mathcal C$ is $n$-ambidextrous?
 A: There is a reference in my comment above (Lemma 2.1.5 in arxiv.org/pdf/1811.02057.pdf). Let me repeat the argument from there in the answer here. 
Let $X$ be a $\pi$-finite space and $q:X\to pt$ be the projection to the point. We assume that the norm map $Nm_q : q_! \to q_*$ is an isomorphism when evaluated at "trivial modules", which are just functors of the form $q^*A$. We want to prove that it is an isomorphism at every object. 
For an object $A$, the zigzag indentity gives us a retract diagram $q_*A \to q_*q^*q_*A \to q_*A$. Now, the norm map $Nm: q_! q^*q_*A \to q_*q^*q_*A$ is an isomorphism by the assumption, and the naturality of the norm map gives us that the composition 
$q_*A\to q_*q^*q_*A \stackrel{Nm^{-1}}{\to} q_!q^*q_*A \to q_!A \stackrel{Nm}{\to}q_*A$ is the identity (where the unlabeled arrows are the usual units and counits). Hence, $Nm: q_! A \to q_*A$ is invertible from the left. A similar dual argument with  the Zigzag identity for $q_!$ and $q^*$ gives us that it is right invertible, hence an isomorphism. 
The intuition here is that taking homotopy orbits/fixed points twice just enlarge the object so this way we reduce from general module to a trivial one.      
