I will try to make it as simple and intuitive as possible.
In the Schrödinger picture, the expectation value of a given operator $\hat{\xi}$ (which itself is frozen in time) is defined as follows (with $\psi(t)$ the wavefunction of our system at time $t$):
$$\langle \hat{\xi} (t) \rangle = \langle \psi (t) \lvert \hat{\xi} \rvert \psi(t) \rangle$$
Which is just the average value of the observable corresponding to $\hat{\xi}$ if a measurement is made at time $t.$ Now exactly because the expectation value creates a direct link between what we predict with our theory in QM with what we observe experimentally, then logically however one goes about defining quantum mechanics, we should obtain the same values for $\langle \hat{\xi} (t) \rangle$ to ensure that we're going to predict the correct experimentally expected values (and hence be able to claim then that the two pictures are equivalent).
To show this equivalence, we first use an important property of the unitary time evolution operator, namely
$$\psi(t_1) = \hat{U}(t_1,t_0) \psi(t_0)$$
i.e. we propagate our wavefunction in time be acting $\hat{U}$ on it. With this, we can now redefine the wavefunction at time $t$ as its value at time $t=0$ upon which we act $\hat{U}(t,0).$ So we rewrite (by a simple substitution) our original expression for $\langle \hat{\xi} (t) \rangle$ as:
$$
\langle \psi (t) \lvert \hat{\xi} \rvert \psi(t) \rangle = \langle \psi (0) \lvert \hat{U}^{\dagger}(t,0) \hat{\xi} \hat{U}(t,0)\rvert \psi(0) \rangle
$$
From the above you can already see the freedom of choice, i.e. to decide to act the time operators either on the wavefunctions or on the operator, by choosing the latter we get:
$$
\langle \psi (0) \lvert \left(\hat{U}^{\dagger}(t,0) \hat{\xi} \hat{U}(t,0)\right)\rvert \psi(0) \rangle = \langle \psi (0) \lvert \hat{\xi}(t) \rvert \psi(0) \rangle
$$
Hence we have successfully shown that the time dependence can also be implemented in the operators, instead of wavefunctions while obtaining the same expectation values for our chosen observable, so let's call $\psi(0) = \psi_h$ with $h$ for Heisenberg, and similarly $\hat{\xi}(t) = \hat{\xi}_h(t).$ With this notation then you can easily relate the operator in the Schrödinger picture with that of the Heisenberg picture by:
$$\hat{\xi}_h(t)=\hat{U}^{\dagger}(t,0) \hat{\xi}_{\rm Schrödinger} \hat{U}(t,0)$$
Finally, from here you can straightforwardly obtain the expression of Heisenberg's equation of motion (although you didn't ask for it, but we've come all this way, may as well show it...):
Take the time derivative of $\hat{\xi}_h(t)$ (using the last equation derived) and by using the relation $d\hat{U}/dt=-\frac{i}{\hbar}\hat{H}\hat{U}$ (and also that $[\hat{H},\hat{U}]=0$):
$$
\begin{align*}
\frac{d\hat{\xi}_h (t)}{dt} &= \frac{d\hat{U}^{\dagger}}{dt} \hat{\xi} \hat{U} + \hat{U}^{\dagger} \hat{\xi}\frac{d\hat{U}}{dt} \\
&= \frac{-1}{i\hbar}(\hat{U}^{\dagger}\hat{H}\hat{\xi}\hat{U}-\hat{U}^{\dagger}\hat{\xi}\hat{H}\hat{U})\\
&=\frac{1}{i\hbar}[\hat{\xi}_h (t),\hat{H}].
\end{align*}
$$
