Two isomorphic finite-dimensional $C^*$-algebras with infinite eigenspaces Let $H_1$, $H_2$ be two separable Hilbert spaces. Let $\mathscr{A}_1$, $\mathscr{A}_2$ be two isomorphic finite-dimensional $C^*$-algebras of operators acting on $H_1$, $H_2$ respectively. Suppose that any eigenvalue of any operator from these algebras has infinite multiplicity. (Suppose also that the identity operators belong to the algebras.)
Can we say that there is a unitary $\mathcal{U}:H_1\to H_2$ such that $\mathfrak{n}(\mathcal{A})=\mathcal{U}\mathcal{A}\mathcal{U}^{-1}$ for any $\mathcal{A}\in\mathscr{A}_1$, where $\mathfrak{n}:\mathscr{A}_1\to\mathscr{A}_2$ is a $*$-isomorphism?
The question was inspired by the comments to Two isomorphic $C^*$-algebras. What is the isomorphism between corresponding Hilbert spaces?
 A: If you allow the units of the algebras to not be the unit of $B(H_j)$, then the answer is no. 
So assume that both subalgebras are unital in their respective $B(H_j)$. Since $\mathscr A_j$ is finite-dimensional, we have 
$$
\mathscr A_j=\bigoplus_{n=1}^m M_{k(n)}(\mathbb C),\ \ \ j=1,2.
$$
Consider the canonical matrix units for these algebras: we have$\{ E_{st}^{n}\}$, $n=1,\ldots,m$, $s,t=1,\ldots,k(n)$ matrix units for $\mathscr A_1$, with $\mathscr A_1=\operatorname{span}\{E_{st}^n:\ n,s,t\}$ and the usual matrix unit relations
$$\tag1
E_{st}^nE_{ab}^p=\delta_{np}\,\delta_{ta}\,E_{sb}^n,\ \ \ \ (E_{st}^n)^*=E_{ts}^n. 
$$
Also, $\sum_{j=1}^{k(n)}E_{ss}^n=I_{k(n)}$, and $\mathscr A_1=\operatorname{span}\,\{E^n_{st}:\ n,s,t\}$. For $\mathscr A_2$ Let $F^n_{st}=\mathfrak{n}(E_{st}^n)$; it is clear that we have $\{F_{st}^n\}$ with the exact same properties as $(1)$. 
The hypothesis is that $E_{ss}^nH_1$ is infinite-dimensional, and similarly $F_{ss}^nH_2$ for all $n$ and all $s$. Now construct, for each $n$ and $s$, orthonormal bases $\{e^{n,s}_\alpha\}$ of $E_{ss}^nH_1$ and $\{f^{n,s}_\alpha\}$ of $F_{ss}^nH_2$, in the following way: fix $n$; let $\{e_\alpha^{n,1}\}_\alpha$ be an orthonormal basis of $E^n_{11}H_1$. Define, for $s>1$, $e_\alpha^{n,s}=E^n_{s1}e_\alpha^{n,1}$; since $E^n_{s,1}$ is an isometry from $E^n_{11}H_1$ to $E^n_{ss}H_1$, it follows that $\{e_\alpha^{n,s}\}_\alpha$ is an orthonormal basis of $E^n_{ss}H_1$. Now use the same idea to produce $\{f_\alpha^{n,s}\}_\alpha$.
As $I_{H_1}=\sum_n\sum_s E^n_{ss}$, we get that $\{e_\alpha^{n,s}\}_{\alpha,n,s}$ is an orthonormal basis of $H_1$. Similarly with $\{f_\alpha^{n,s}\}_{\alpha,n,s}$ and $H_2$. 
Define a unitary $V:H_1\to H_2$ by $Ve^{n,s}_\alpha=f^{n,s}_\alpha$. 
Now
\begin{align}
VE_{st}^n e^{m,v}_\alpha&=VE_{st}^nE_{v1}^me_\alpha^{m,1}=\delta_{nm}\,\delta_{vt}\,VE_{s1}^ne_\alpha^{n,1}=\delta_{nm}\,\delta_{vt}\,Ve_\alpha^{n,s}
=\delta_{nm}\,\delta_{vt}\,f_\alpha^{n,s}\\ \ \\
&=\delta_{nm}\,\delta_{vt}\,F^n_{s1}f_\alpha^{n,1}
=F^n_{st}F^m_{v1}f_\alpha^{m,1}
=F^n_{st}f_\alpha^{m,v}
=F^n_{st}Ve_\alpha^{m,v}.
\end{align}
Thus $VE_{st}^n=F_{st}^nV$, and then $F_{st}^n=VE_{st}^nV^*$. That is, 
$$\tag2
\mathfrak n(E_{st}^n)=VE_{st}^nV^*. 
$$
As the matrix units span $\mathscr A_1$, the equality $(2)$ holds for all $T\in\mathscr A_1$. 
