Is it true that $V$ and $V^*$ are naturally isomorphic as finite vector spaces if $V$ is equipped with an inner product?

Question

This is a homework question from my differentiable manifolds class:

In general we know that if $\dim V<\infty$ then $V$ and $V^*$ are isomorphic because any two vector spaces with the same dimension are isomorphic. But the fact that any two vector spaces with the same dimension are isomorphic depends on our choice of bases for them, therefore this isomorphism is not natural in general.

Now suppose that $V$ is an inner product space. So, there's a non-degenerate bilinar form on $V$ like $<\cdot,\cdot> :V\times V\to\mathbb{R}$. Can $V$ and $V^*$ be shown to be naturally isomorphic now?

Here's my thought:

Now that $V$ has been equipped with an inner product we have this natural pairing $T$ that gives us our desired coordinate-free isomorphism:

$T: V \to V^*$

$ v \mapsto f_v(x) = <v,x>$

1. $\forall \lambda \in \mathbb{R}, \forall v_1,v_2 \in V:$

   $$\forall x \in V: f_{v_1+\lambda v_2}(x) = <v_1+\lambda v_2,x> = <v_1,x>+\lambda <v_2,x>=f_{v_1}(x)+\lambda f_{v_2}(x)$$

Therefore $T(v_1+\lambda v_2)=f_{v_1+\lambda v_2}=f_{v_1}+\lambda f_{v_2}=T(v_1)+\lambda T(v_2)$ and $T$ is a linear transformation.

1. $\ker{T}=\{ v: T(v)=0 \}=\{v: f_v(x)=<v,x>=0 \text{ forall x in V}\}=\{0\} $ because $<\cdot,\cdot>$ has been assumed to be non-degenerate. Therefore $T$ is one-to-one.

2. $\operatorname{Image}{T}=V^*$ because of rank-nullity theorem and the fact that $\dim{V^*}=\dim{V}$. More precisely, $\dim{V} = \ker{T} + \operatorname{rank}{T} = 0 + \operatorname{rank}{T} \implies \operatorname{rank}{T} = n \implies \operatorname{Image}{T}=V^*$

Therefore $T$ is a one-to-one surjective linear transformation that has been defined without reference to any basis for $V$.

Now how can I show that this construction is really a natural isomorphism in the category of finite vector spaces? I mean, assuming that my construction is correct, how can I show that it fits the definition of a natural transformation in category theory?

This question has a sequel that asks me to prove several isomorphisms about the vector space $\Lambda^k(V)$ like: $\Lambda^k(V) \cong \Lambda^{n-k}(V)$. But first I need to understand this one carefully I think.