Let $A$ and $B$ be two finite-dimensional C$^\ast$-algebras which are algebraically isomorphic, i.e. there exists a bijective linear map $\Phi:A\rightarrow B$ such that $\Phi(ab)=\Phi(a)\Phi(b)$ for all $a,b \in A$. Is it then true that $\Phi$ is already a $\ast$-isomorphism in the sense that $\Phi$ is isometric and $\Phi(a^\ast)=\Phi(a)^\ast$ for every $a \in A$?
I guess – since injective $\ast$-homomorphism are isometric – it suffices to show that $\Phi$ preserves the involution. For this again it suffices to show that self-adjoint elements are mapped to self-adjoint elements. But how to proceed?
Best Answer
The title of your question and the body of your question are actually two distinct questions. In what follows, I'll appeal to facts about finite-dimensional $C^\ast$-algebras that you can find, for instance, in D. Farenick’s Algebras of Linear Transformations, though I don't guarantee that my answer is absolutely optimal in presentation.
The title of your question asks the following: is it true that two finite-dimensional $C^\ast$-algebras $A$ and $B$ are isomorphic as unital associative algebras over $\mathbb{C}$ if and only if they are isometrically $\ast$-isomorphic? The answer to this question is yes.
Recall Wedderburn's theorem for finite-dimensional semisimple $\mathbb{C}$-algebras:
Now, not only is every finite-dimensional $C^\ast$-algebra $A$ semisimple in the above sense, but in this case the isomorphism $A \to \bigoplus_{i=1}^k M_{n_i}(\mathbb{C})$ in Wedderburn's theorem can always be chosen to be a $\ast$-isomorphism, which is therefore bicontinuous (by finite-dimensionality of domain and range) and hence isometric (as a bicontinuous $\ast$-isomorphism between $C^\ast$-algebras).
So, suppose that $A$ and $B$ are finite-dimensional $C^\ast$-algebras that are isomorphic as unital associative algebras over $\mathbb{C}$. By the $C^\ast$-algebraic refinement of Wedderburn's theorem, there exist unique finite non-decreasing lists $(m_1,\dotsc,m_d)$ and $(n_1,\dotsc,n_k)$ of natural numbers, such that $A \cong \bigoplus_{i=1}^d M_{m_i}(\mathbb{C})$ and $B \cong \bigoplus_{j=1}^k M_{n_j}(\mathbb{C})$ as unital $C^\ast$-algebras; fix isometric $\ast$-isomorphisms $\phi_A : A \to \bigoplus_{i=1}^d M_{m_i}(\mathbb{C})$ and $\phi_B : B \to \bigoplus_{j=1}^k M_{n_j}(\mathbb{C})$. But now, since $A \cong B$ as unital associative algebras over $\mathbb{C}$, so too are $\bigoplus_{i=1}^d M_{m_i}(\mathbb{C}) \cong \bigoplus_{j=1}^k M_{n_j}(\mathbb{C})$ as unital associative algebras over $\mathbb{C}$, so that by Wedderburn's theorem itself, we must have $(m_1,\dotsc,m_d) = (n_1,\dotsc,n_k)$. Thus, the isometric $\ast$-isomorphisms $\phi_A$ and $\phi_B$ have the same range, so that $\phi_B^{-1} \circ \phi_A : A \to B$ is the desired isometric $\ast$-isomorphism.
Now, the body of your question asks the following: given finite-dimensional $C^\ast$-algebras $A$ and $B$, would an isomorphism $\phi : A \to B$ of unital associative algebras necessarily an isometric $\ast$-isomorphism? The answer to this is no precisely because there exist unital associative $\mathbb{C}$-algebra automorphisms of finite-dimensional $C^\ast$-algebras that are not $\ast$-automorphisms.
For example, let $S = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$, and define a unital associative $\mathbb{C}$-algebra automorphism of $M_2(\mathbb{C})$ by $$\forall a \in M_2(\mathbb{C}), \quad \phi(a) := SaS^{-1}.$$ Then $$ \phi(S)^\ast = (S S S^{-1})^\ast = S^\ast = \begin{pmatrix}1&0\\1&1\end{pmatrix}, $$ while $$ \phi(S^\ast) = \begin{pmatrix}1&1\\0&1\end{pmatrix} \begin{pmatrix}1&0\\1&1\end{pmatrix} \begin{pmatrix}1&-1\\0&1\end{pmatrix} = \begin{pmatrix}2&-1\\1&0\end{pmatrix} \neq \phi(S)^\ast. $$