$$ \DeclareMathOperator{\dif}{d \!} \newcommand{\ramuno}{\mathrm{i}} \newcommand{\exponent}{\mathrm{e}} \newcommand{\ket}[1]{|{#1}\rangle} \newcommand{\bra}[1]{\langle{#1}|} \newcommand{\braket}[2]{\langle{#1}|{#2}\rangle} \newcommand{\bracket}[3]{\langle{#1}|{#2}|{#3}\rangle} \newcommand{\linop}[1]{\hat{#1}} \newcommand{\dpd}[2]{\frac{\partial #1}{\partial #2}} \newcommand{\dod}[2]{\frac{\dif{#1}}{\dif{#2}}} $$
Using the Schrödinger equation and the definition of the expectation value it can be shown that the time dependence of the expectation value of an observable $A$ for a system in arbitrary state $\ket{\Psi(t)}$ is given by $$ \dod{\langle A \rangle}{t} = \frac{\ramuno}{\hbar} \langle [\linop{H}, \linop{A}] \rangle + \big\langle \dpd{\linop{A}}{t} \big\rangle \, , \tag{1} $$ and this equation shows that, in general, if an operator $\linop{A}$ commutes with the Hamiltonian operator $\linop{H}$ and does not have an explicit time dependence, then the expectation value of the corresponding observable $A$ is time independent.
For stationary states $\ket{\Psi(t)} = \exponent^{-\ramuno E_{k} t / \hbar} \ket{E_{k}}$ the first term in the expression for the time dependence of the expectation value of an observable vanishes $$ \langle [\linop{H}, \linop{A}] \rangle = \bracket{ \Psi(t) }{ \linop{H} \linop{A} }{ \Psi(t) } - \bracket{ \Psi(t) }{ \linop{A} \linop{H} }{ \Psi(t) } = \bracket{ E_{k} }{ \linop{H} \linop{A} }{ E_{k} } - \bracket{ E_{k} }{ \linop{A} \linop{H} }{ E_{k} } = E_{k} \bracket{ E_{k} }{ \linop{A} }{ E_{k} } - E_{k} \bracket{ E_{k} }{ \linop{A} }{ E_{k} } = 0 \, , $$ and so the time dependence of the expectation value is simply given by $$ \dod{\langle A \rangle}{t} = \big\langle \dpd{\linop{A}}{t} \big\rangle \, . \tag{2} $$
Nevertheless, the statement like the following one
A stationary state is called stationary because the system remains in the same state as time elapses, in every observable way. Wikipedia
is found in many books.
The thing that troubled me is the word every, since from (2) it appears that if an operator $\linop{A}$ carries some explicit time dependence, then the expectation value of the corresponding observable $A$ changes in time. So stationary states are, in fact, not so stationary.
I have the feeling that I am missing something. And our discussion with Bubble helped clarify what's bothering me.
As far as I know operators in the Schrödinger picture usually do not carry an explicit time dependence. Again, usually, but not always. In many books (see, for instance, Griffiths, D.J., Introduction to quantum mechanics, 2nd ed.) one can find that
Operators that depend explicitly on $t$ are quite rare, so almost always $\dpd{Q}{t} = 0$.
And, yet the author claims that
Every expectation value is constant in time.
I feel like there is a gap between operators being almost always explicitly independent of time and every expectation value being constant.