= Solution
Interpret the involution as the adjoint for the Hilbert-space structure on $V$. The hypotheses make $\Omega$ a <cyclic vector for an operator algebra> for both $A$ and $A'$. They also make it a <separating vector> for each: if $a\Omega=0$, then $a$ vanishes on $A'\Omega=V$, and similarly for $A'$. Thus the <Tomita operator>
$$
S(a\Omega)=a^*\Omega
$$
is well defined, antilinear, invertible and satisfies $S^2=I$. Define the <modular operator> and <modular conjugation> by
$$
\boxed{\Delta=S^*S,\qquad J=S\Delta^{-1/2},\qquad S=J\Delta^{1/2}.}
$$
We calculate them explicitly to prove the commutant assertion.
The finite-dimensional structure of a <unital> <star-algebra> gives an orthogonal decomposition
$$
V=\bigoplus_j(\mathbb C^{n_j}\otimes\mathbb C^{m_j}),\qquad
A=\bigoplus_j(M_{n_j}(\mathbb C)\otimes I_{m_j}),\qquad
A'=\bigoplus_j(I_{n_j}\otimes M_{m_j}(\mathbb C)).
$$
This is the usual matrix-block structure: minimal central <projections> separate the simple summands, and each simple <matrix algebra> acts as its defining representation with a multiplicity space. Identify each <vector> block with an $n_j$-by-$m_j$ <matrix> $D_j$ with its <Hilbert-Schmidt> <inner product>. The two algebras act by left and right multiplication. If $D_j$ has rank $r_j$, the left orbit has dimension $n_jr_j$ and the right orbit dimension $m_jr_j$. Cyclicity on both sides therefore forces $r_j=m_j=n_j$. Every $D_j$ is square and invertible.
Right multiplication by a suitable unitary is a unitary change of this identification commuting with the left algebra. The polar decomposition of each $D_j$ therefore lets us assume $D_j>0$. In one such block, for an arbitrary <matrix> $X=aD$, the <Tomita operator> is
$$
S(X)=D^{-1}X^*D.
$$
Its polar factors are
$$
\boxed{J(X)=X^*,\qquad \Delta^{1/2}(X)=DXD^{-1},\qquad
\Delta(X)=D^2XD^{-2}.}
$$
Indeed, $J$ is an antiunitary involution and $J\Delta^{1/2}(X)=D^{-1}X^*D$. The fact that $\Delta^{1/2}$ is a <positive operator> is transparent in a <basis> diagonalizing $D$: it multiplies the <matrix> unit $E_{kl}$ by the positive ratio $d_k/d_l$. These are therefore the unique polar factors of $S$.
Finally, if $L_a$ denotes left multiplication,
$$
(JL_aJ)(X)=(aX^*)^*=Xa^*.
$$
As $a$ ranges over the full <matrix> block, these are exactly its right multiplications. Applying this on every block and transporting back by the unitary identifications proves
$$
\boxed{JAJ=A'.}
$$
This also establishes the required modular definitions intrinsically, independent of the chosen <matrix> coordinates.
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