= Solution
Put $c=\cos(\theta/2)$ and $s=\sin(\theta/2)$. The detector rotation is real and orthogonal, so its inverse is the same transformation. Substituting it gives
$$
|\Phi^C\rangle=\frac1{\sqrt2}\left[
(c|\uparrow\rangle+s|\downarrow\rangle)|d_\nearrow\rangle
+(s|\uparrow\rangle-c|\downarrow\rangle)|d_\swarrow\rangle\right].
$$
Thus the requested orthonormal system states are
$$
\boxed{|\nearrow\rangle=c|\uparrow\rangle+s|\downarrow\rangle,\qquad
|\swarrow\rangle=s|\uparrow\rangle-c|\downarrow\rangle.}
$$
The pure <density operator> has not changed: this is only a different <Schmidt decomposition> of a maximally entangled state. But choosing a nonselective <projective measurement> in this new detector basis gives a different operation and generally a different state:
$$
\boxed{\rho_\theta=\frac12\left(|\nearrow d_\nearrow\rangle\langle\nearrow d_\nearrow|
+|\swarrow d_\swarrow\rangle\langle\swarrow d_\swarrow|\right).}
$$
For an explicit matrix in the original ordered basis, write $q=c^2s^2$ and $h=cs(c^2-s^2)/2$. Then
$$
\rho_\theta=\begin{pmatrix}
\frac12-q&h&h&q\\h&q&q&-h\\h&q&q&-h\\q&-h&-h&\frac12-q
\end{pmatrix}.
$$
At $\theta=\pi/2$ the entries $(1,4)$ and $(2,3)$ are $1/4$, whereas they vanish in $\rho_{\rm up/down}$. Nevertheless, the system and detector individually each have <reduced density matrix> $I/2$ for either mixture.
This is the <preferred-basis problem>: the correlated state alone does not specify which detector observable becomes a stable classical record. Arbitrarily declaring collapse in a Schmidt basis changes the physical prediction. The actual detector interaction and environment must supply the basis selection, rather than a mere choice of notation.
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