Thinking about your question has led me to the following conclusion:
Suppose there is a a conserved quantity $Y$, and an isolated system $S$. Let's also suppose the isolation is temporarily lifted to measure an observable $\widehat{X}$ of $S$. This measurement is performed by a measuring apparatus $A$. For simplicity, let's lump together the rest of the universe outside the system including the apparatus and the environment into $A$. Let's further assume we have a tensor product structure between $S$ and $A$ for the Hilbert space. Let $\widehat{Y}_S$ and $\widehat{Y}_A$ be the restriction of $\widehat{Y}$ to $S$ and $A$ respectively. Then, the sum $\widehat{Y}_S + \widehat{Y}_A$ has to be conserved.
Now, let's suppose $\widehat{Y}_S$ and $\widehat{X}$ don't commute. Let's see what happens if we further make an assumption very common in measurement theory that for systems starting out in an eigenstate of $X$, a perfect measurement will leave it in an eigenstate of $X$ with the same eigenvalue. The Hilbert space of $A$ decomposes into eigenspaces of a pointer operator $\widehat{P}$ such that after a measurement of an eigenstate of $X$, $A$ ends up in an eigenstate of $\widehat{P}$ with an eigenvalue equal to the eigenvalue of $X$.
As $\widehat{Y}_S$ and $\widehat{X}$ don't commute, there exists an eigenvalue of $X$ such that all nonzero eigenstates of it aren't also $\widehat{Y}_S$ eigenstates. With all these assumptions in place, an initial state of $\sum_n c_n \left| x_n \right\rangle$ will evolve into
$$\sum_n c_n \left| x_n \right\rangle \otimes \left| \chi_n \right\rangle$$
where $\left| \chi_m \right\rangle$ and $\left| \chi_n \right\rangle$ are orthogonal if $m \neq n$ because they have different pointer values. Let's further assume $A$ is initially in an eigenstate of $\widehat{Y}_A$. Assume that the system is initially in an eigenstate of $\widehat{Y}_S$ with eigenvalue $y_m$ which isn't also an eigenstate of $X$. Then, this state can be expressed as $\sum_n U_{mn} \left| x_n \right\rangle$ with $U_{mn}$ being nonzero for at least two different values of $n$. The final state is
$$\sum_{np} U_{mn} U_{pn}^* \left| y_p \right\rangle \otimes \left| \chi_n \right\rangle$$
If we look at the contribution to the sum for one of the values of $n$ for which $U_{mn}$ is nonzero, we find the combined $S+A$ system can't be in an eigenstate of $\widehat{Y}$. Precisely because the $\chi$'s are orthogonal, the contributions from different $n$'s can't cancel, and this remains true for the wavefunction as a whole.
But of course, $\widehat{Y}$ generates a symmetry, and $\widehat{X}$ is not invariant under this symmetry. So, to measure $X$, the initial state of the apparatus can't possibly be invariant under this symmetry either. So, we shouldn't expect $A$ to start in an eigenstate of $\widehat{Y}_A$. The measurement process itself is expected to change $\widehat{Y}_S$, but we have a compensating change in $\widehat{Y}_A$.