A simple example:
Take a real wave function $\psi(x)$ in position space, normalized as $$\int\limits_{-\infty}^{+\infty} |\psi(x)|^2=1.$$ $\psi(x)$ represents a certain (pure) state of the system with the probabilty distribution $|\psi(x)|^2=\psi(x)^2$ for measurements of the observable "position of the particle" in this state.
Compare this with the state represented by the wave function $\chi(x)=\psi(x)e^{ip_0x/\hbar}$, where $p_0$ is a real constant (with the dimension of a momentum). As $|e^{i p_0 x/\hbar}|=1$, the state $\chi$ has the same probability distribution $|\chi(x)|^2=|\psi(x)|^2$ as the state $\psi$ for position measurents. In other words, the states $\psi$ and $\chi$ cannot be distinguished by measurements of the observable "position".
But what happens, if we measure the observable "momentum", represented by the differential operator $P=-i \hbar d/dx$, in these two states? Let us simply have a look at the expectation value of the momentum (the mean value of the momentum) in these two states. According to the rules of quantum mechanics, the expectation value of the momentum in the state $\psi$ (remember, $\psi(x)$ was assumed to be real) is given by $$\begin{align} \langle \psi|P \psi\rangle &= \int\limits_{-\infty}^{+\infty} \!\!\!dx\, \psi(x)\left(-i \hbar \frac{d}{dx}\psi(x) \right) \\ &=\frac{-i \hbar}{2}\int\limits_{-\infty}^{+\infty}\!\!\! dx \, \frac{d}{dx} \psi(x)^2 \\ &= \frac{-i \hbar}{2}\left[\psi(+\infty)^2-\psi(-\infty)^2 \right]=0, \end{align}$$ as $\lim\limits_{x\to \pm \infty}\psi(x)=0.$
The physical interpretation of this result is that measuremts of the momentum with outcomes $p_1, p_2,\ldots,p_N$ performed at a large number $N$ of copies of the system prepared in the state $\psi$ will have the property $$\lim\limits_{N\to\infty} \frac{1}{N}\sum\limits_{k=1}^N p_k =0.$$
What happens in the case of a momentum measurement if the system is prepared in the state $\chi$? The expectation value of the momentum is now given by $$\begin{align}\langle \chi |P \chi\rangle &= \int\limits_{-\infty}^{+\infty} \!\!\! dx \,\chi^\ast(x) \left(-i \hbar\frac{d}{dx} \right) \chi(x) \\ &= \int\limits_{-\infty}^{+\infty} \!\!\! dx \, e^{-ip_0x/\hbar} \psi(x) \left( -i \hbar \frac{d}{dx} \right) \left(\psi(x) e^{ip_0x/\hbar} \right) \\ &=-i\hbar \int\limits_{-\infty}^{+\infty} \!\!\! dx \, \psi(x) \left( \psi^\prime(x)+\frac{i p_0}{\hbar} \psi(x) \right)\\ &=p_0, \end{align}$$ showing that the experimentally measured values $p_1, p_2,\ldots, p_N$ of the momentum are now distributed in such a way that $$\lim\limits_{N\to \infty} \frac{1}{N}\sum\limits_{k=1}^N p_k = p_0.$$
The probability distribution of the momentum of a particle in an arbitrary state $\phi$ can be studied in full detail by taking advantage of fact that the momentum-space wave function $\tilde{\phi}(p)$ is related to the associated wave function $\phi(x)$ in position space by the Fourier transform $$\tilde{\phi}(p)=\int\limits_{-\infty}^{\infty}\!\!\! dx \, \frac{e^{-ipx/\hbar}}{\sqrt{2\pi \hbar}} \phi(x),$$ where the probabilty distribution for momentum measurements is given by $|\tilde{\phi}(p)|^2$. As the position-space wave function $\phi(x)$ can be recovered from the momentum-space wave function $\tilde{\phi}(p)$ by the inverse Fourier transform $$\phi(x) = \int\limits_{-\infty}^{+\infty} \!\!\! dp \, \frac{e^{ipx/\hbar}}{\sqrt{2 \pi \hbar}} \tilde{\phi}(p),$$ both, $\phi(x)$ and $\tilde{\phi}(p)$, contain the same information about the state $\phi$.
Returning to our previous example, the momentum-space wave functions of the states $\psi$ and $\chi$ are thus related by $$\tilde{\chi}(p) =\int\limits_{-\infty}^{+\infty} \!\!\! dx \frac{e^{-ipx/\hbar}}{\sqrt{2\pi \hbar}} \psi(x) e^{ip_0x/\hbar} = \int\limits_{-\infty}^{+\infty} \!\! \! dx \frac{e^{-i(p-p_0)x/\hbar}}{\sqrt{2 \pi \hbar}} = \tilde{\psi}(p-p_0),$$ showing that the probability distribution $|\tilde{\chi}(p)|^2=|\tilde{\psi}(p-p_0)|^2$ of the momenta in the state $\chi$ is related to the probability distribution $|\tilde{\psi}(p)|^2=|\tilde{\psi}(-p)|^2$ in the state $\psi$ by a translation in momentum space, explaining the above outcomes for the expectation values of the momentum.
In conclusion, the states $\psi$ and $\chi$ have the same probability distribution for the observable "position", but their probability distributions for the momentum differ.
Note that $\psi(x) \to e^{i \alpha} \psi(x)$ with a constant phase angle $\alpha \in \mathbb{R}$ does not change the physics. Pure states are described by "rays" in Hilbert space.
Finally, a short remark on the general case. Assume, you decompose a state vector $|\psi\rangle$ with respect to the complete orthonormal system $\{|\phi_n\rangle\}_{n=1}^\infty$ of eigenvectors of some observable $A=\sum\limits |\phi_n \rangle a_n \langle \phi_n|$ with nondegenerate real eigenvalues $a_1, a_2, \ldots$ by $$|\psi\rangle = \sum\limits_{n=1}^\infty c_n |\phi_n\rangle, \qquad \sum\limits_{n=1}^\infty |c_n|^2 =1,$$ where $|c_n|^2$ is the probability to measure the eigenvalue $a_n$ of $A$ in this state. A change of the expansion coefficients $c_n \to c_n e^{i \alpha_n} $ with $\alpha_n \in \mathbb{R}$ will not change the probability distribution for the eigenvalues $a_n$ of the observable $A$ in the new state $$|\chi\rangle =\sum\limits_{n=1^\infty} c_n e^{i \alpha_n} |\phi_n\rangle,$$ but the probability distribution for the eigenvalues $b_k$ of some observable $B$ with $[A,B]\ne 0$ will change (unless $\alpha_n =\alpha \, \forall \, n$).
