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Amir Sagiv
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If $G$ is an affine algebraic group (for example a finite group), then the category of k$k$-linear cocontinuous symmetric monoidal functors from $\mathsf{Rep}(G)$ to $\mathsf{Vect}_k$ is equivalent to the category of $G$-torsors over $k$. In particular, not every such functor needs to be isomorphic to the identity. For example, if $k'$ is finite Galois extension of k with Galois group $G$, then the functor $F(V) = (V \otimes_{k} k')^{G}$ will satisfy all the axioms you will think to write down, but is not isomorphic to the identity functor.

If $G$ is an affine algebraic group (for example a finite group), then the category of k-linear cocontinuous symmetric monoidal functors from $\mathsf{Rep}(G)$ to $\mathsf{Vect}_k$ is equivalent to the category of $G$-torsors over $k$. In particular, not every such functor needs to be isomorphic to the identity. For example, if $k'$ is finite Galois extension of k with Galois group $G$, then the functor $F(V) = (V \otimes_{k} k')^{G}$ will satisfy all the axioms you will think to write down, but is not isomorphic to the identity functor.

If $G$ is an affine algebraic group (for example a finite group), then the category of $k$-linear cocontinuous symmetric monoidal functors from $\mathsf{Rep}(G)$ to $\mathsf{Vect}_k$ is equivalent to the category of $G$-torsors over $k$. In particular, not every such functor needs to be isomorphic to the identity. For example, if $k'$ is finite Galois extension of k with Galois group $G$, then the functor $F(V) = (V \otimes_{k} k')^{G}$ will satisfy all the axioms you will think to write down, but is not isomorphic to the identity functor.

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If G$G$ is an affine algebraic group (for example a finite group), then the category of k-linear cocontinuous symmetric monoidal functors from Rep(G)$\mathsf{Rep}(G)$ to Vect_k$\mathsf{Vect}_k$ is equivalent to the category of G$G$-torsors over k$k$. In particular, not every such functor needs to be isomorphic to the identity. For example, if k'$k'$ is finite Galois extension of k with Galois group G$G$, then the functor $F(V) = (V \otimes_{k} k')^{G}$ will satisfy all the axioms you will think to write down, but is not isomorphic to the identity functor.

If G is an affine algebraic group (for example a finite group), then the category of k-linear cocontinuous symmetric monoidal functors from Rep(G) to Vect_k is equivalent to the category of G-torsors over k. In particular, not every such functor needs to be isomorphic to the identity. For example, if k' is finite Galois extension of k with Galois group G, then the functor $F(V) = (V \otimes_{k} k')^{G}$ will satisfy all the axioms you will think to write down, but is not isomorphic to the identity functor.

If $G$ is an affine algebraic group (for example a finite group), then the category of k-linear cocontinuous symmetric monoidal functors from $\mathsf{Rep}(G)$ to $\mathsf{Vect}_k$ is equivalent to the category of $G$-torsors over $k$. In particular, not every such functor needs to be isomorphic to the identity. For example, if $k'$ is finite Galois extension of k with Galois group $G$, then the functor $F(V) = (V \otimes_{k} k')^{G}$ will satisfy all the axioms you will think to write down, but is not isomorphic to the identity functor.

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Jacob Lurie
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Oops.If G is an affine algebraic group (for example a finite group), then the category of k-linear cocontinuous symmetric monoidal functors from Rep(G) to Vect_k is equivalent to the category of G-torsors over k. In particular, not every such functor needs to be isomorphic to the identity. misunderstoodFor example, if k' is finite Galois extension of k with Galois group G, then the questionfunctor $F(V) = (V \otimes_{k} k')^{G}$ will satisfy all the axioms you will think to write down, but is not isomorphic to the identity functor.

Oops... misunderstood the question.

If G is an affine algebraic group (for example a finite group), then the category of k-linear cocontinuous symmetric monoidal functors from Rep(G) to Vect_k is equivalent to the category of G-torsors over k. In particular, not every such functor needs to be isomorphic to the identity. For example, if k' is finite Galois extension of k with Galois group G, then the functor $F(V) = (V \otimes_{k} k')^{G}$ will satisfy all the axioms you will think to write down, but is not isomorphic to the identity functor.

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Jacob Lurie
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