Algebraic isomorphism between two C$^\ast$-algebras implies $\ast$-isomorphic 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?
 A: 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:

Let $A$ be a finite-dimensional unital associative algebra over $\mathbb{C}$ that is semisimple (i.e., is semisimple as a left module over itself).
There exists a finite list of natural numbers $(n_1,\dotsc,n_k)$, unique up to reordering, such that $A \cong \bigoplus_{i=1}^k M_{n_i}(\mathbb{C})$ as unital associative 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
$$
 \phi(S^\ast) = \begin{pmatrix}1&1\\0&1\end{pmatrix} \begin{pmatrix}1&0\\1&1\end{pmatrix} \begin{pmatrix}1&-1\\0&1\end{pmatrix} = \begin{pmatrix}2&-1\\1&0\end{pmatrix} \neq \phi(S)^\ast.
$$
