The set of irreducible representations (over C) determines a finite group I'm trying to understand group representations, and I think that this statement is correct (where determines means up to isomorphism), although I couldn't find a proof online (without references to more complicated things like "Tannaka duality"), so I tried to prove it, but I may have overlooked something:
Let $G = \{x_1, ..., x_n\}$, $H = \{y_1, ..., y_n\}$ be two finite groups of same cardinality.
Suppose G and H have the same set of irreducible representations in the following sense :
there is a finite set of morphisms $g_i$ (resp. $h_i$), each one from G (resp. H) to
some subgroup of $GL(n_i, \Bbb{C})$ and we can write $g_i(x_j) = P_i h_i(y_j) P_i^{-1}$ for some invertible matrix $P_i$ and all $j$.
Then if we consider the regular representation of G, we can write it as a sum of the $g_i$ representations
(with $n_i$ giving the multiplicity) and do the same for H.
That shows that the two regular representations are isomorphic as we can build a new base using a block-diagonal transition matrix using $(P_i)$.
As they are also faithful (isomorphic to the groups themselves), then G and H are isomorphic.
Is that correct ?
EDIT: I updated the "same set of irreducible representations" following the comments (that also requires the additional assumption of same cardinality)
 A: Your theorem is correct: we can reconstruct a group's operation from a full set of irreducible representations (interpreted as mere functions from $G$ to $GL(V)$s). And your proof is also correct: we can reconstruct the regular representation (as a function) by taking direct sums of the irreps (with multiplicities equal to their dimensions); then $G$ is isomorphic to its image (under the regular rep) so its operation can be recovered by transporting matrix multiplication back to it.
We can probably phrase this in categorical terms, by considering the category whose objects are $(X,\mathcal{F})$ where $X$ is a set of $\mathcal{F}$ is a family of functions on $X$, and whose morphisms are tuples $(\alpha,\beta,\Phi)$ where $\Phi$ is a family of functions $\phi_f$ indexed by $\mathcal{F}$, and $\alpha:X\to Y,\beta:\mathcal{F}\to\mathcal{G}$ satisfy the diagram $f=\phi_f\circ\beta(f)\circ\alpha$.
Given a group $G$, we can construct such an object as $(G,\mathcal{F})$ with $\mathcal{F}$ a complete set of irreducible representations, then any isomorphism $(G_1,\mathcal{F}_1)\to(G_2,\mathcal{F}_2)$ with $\Phi$ consisting of linear operators can be converted into an isomorphism $G_1\to G_2$.

Remiss to not explain why Tannaka-Krein duality (a) is natural and (b) not that complicated for finite groups. First of all, just as every (finite-dimensional) vector space is (not naturally) isomorphic to its dual, every (finitely-generated) abelian group is (not naturally) isomorphic to its dual, aka its character group. However, the dual of the dual is naturally isomorphic in these cases. Indeed, the Pontryagin dual of the Pontryagin dual of a locally compact abelian group is naturally isomorphic to the original group. This is achieved with the Fourier transform on $\mathbb{R}$ (which is self-dual) or between $S^1$ and $\mathbb{Z}$ (which are dual).
The question, then, is if we can generalize this to nonabelian $G$. The characters of an abelian group are its irreps, and pointwise multiplying characters is equivalent to tensoring them. Can we recover a finite group (or more generally, a compact group, if not locally compact) from its irreducible representation and tensor product? The first issue is that irreps are not closed under tensoring, so (being category theorists as well as represenation theorists) we consider the monoidal category of representations of $G$ with the tensor product operation. This turns out to not be enough, but it is if we equip it with the fiber functor $\Phi$ (forgetful functor to the category of vector spaces).
Then $\mathrm{Aut}(\Phi)\cong G$ (tensor-preserving natural functors). Pick a $v\in V$ and consider evaluation-at-$v$ as a morphism $\mathbb{C}[G]\to V$ given by $x\mapsto xv$, to show $\eta\in\mathrm{Aut}(\Phi)$ satisfies $\eta_V(v)=\eta_{\mathbb{C}[G]}(e)v$, then consider the "comultiplication" morphism $\Delta:\mathbb{C}[G]\to\mathbb{C}[G]\otimes\mathbb{C}[G]$ given by $\Delta(g)=g\otimes g$ to show $\eta_{\mathbb{C}[G]}(e)=g$ for some group element $g$. Then $\eta\leftrightarrow g$ is an isomorphism $\mathrm{Aut}(\Phi)\cong G$.
There is still "duality" present here. All intertwiners intertwine with all $\pi_V(g)$ operators, and conversely the natural transformations which intertwine with all intertwiners are precisely of the form $\eta_V=\pi_V(g)$.
A: I think the "correct" way to deduce that $G$ is isomorphic to $H$, without requiring bijection between the two group is the following.
Define a straightening system $\mathcal{S}$ as a finite collection of vector spaces  $V_0, \ldots, V_n$ over $\mathbb{C}$, equipped with an hermitian product, of dimensions $d_0=1, d_1, \ldots, d_n$, plus the following data:

*

*Numbers $n_{ij}^k$ such that
$$ d_i d_j = \sum_k n_{ij}^k d_k$$

*A permutation $\sigma: \in S_{n+1}$ such that $\sigma(0) = 0$;

*Unitary isomorphisms $T_{ij} : V_i \otimes V_j  \to \oplus_k (V_k) ^{n_{ij}^k} $

*Unitary isomorphisms $\phi_i : V_i \to V_{\sigma(i) }$;

For $W$ a vector space equipped with an hermitian product, let $U(W) $ be the space of unitary automorphisms. Also, for $f: V \to W$ an unitary isomorphism, define
$$E(f) : U(V) \to U(W) $$
Be the conjugation by $f$.
Define the little tannaka group of the straightening system as
$$\tau(\mathcal{S}) := \{ (u_i) \in \prod_i U(V_i) : E(T_{ij}) (u_i \otimes u_j) = \oplus_k n_{ij}^k u_k, E(\phi_i) (u_i) = u_{\sigma(i)} $$
The group structure is given by componentwise multiplication. We have the following
Proposition.  *Let $G$ be a group, and let $\mathcal{S}(G) $ be the straightening system defined in this way:

*

*Vector spaces are the irreducible representations and $V_0$ is the trivial representation, and the hermitian product is the only positive definite invariant by the $G$ action;

*$n_{ij}^k = \langle \chi_i \chi_j, \chi_k \rangle $, where $\chi_s$ is the character of the representation $V_s$;

*The permutation is such that $V_i^* \simeq V_{\sigma(i) }$;

*Isomorphisms $T_{ij}$ are given by the representation isomorphism
$$ V_i \otimes V_j \simeq_G \oplus_k (V_k) ^{n_{ij}^k}$$

*Isomorphism $\phi_i$ is given by the repn isomorphism
$$ V_i^* \simeq V_{\sigma(i) }$$
Then the map
$$ G \to \tau(\mathcal{S}(G) ) $$
given by $g \mapsto (\rho_i(g) ) $ is an isomorphism of groups*.
This is not easy to show, but it is true. Let me remark two facts:

*

*The data of spaces and numbers $n_{ij}^k$ alone does not determine the group structure; one can show that these data are equivalent to give the character table of $G$, which is notoriously not equivalent to give the group;

*The additional data of the isomorphisms are enough to determine the group, and two groups will be isomorphic one to each other iff the two straightening systems are isomorphic. Morally the flexibility you have in the straightening system is the choice of the isomorphism $T_{ij}$; you can multiply this by a constant on each summand $V_k$ of $V_i \otimes V_j$ and it will still be an isomorphism of representations. What the proposition says is that this is the only flexibility you have.

