It is consistent with ZFC that the answer is no, but under the Ultrapower Axiom, the answer is yes, and not just for normal ultrafilters, but for arbitrary countably complete ultrafilters.

**First I'll show that in the Kunen-Paris model, there exist distinct normal ultrafilters $U_0$ and $U_1$ such that $j_{U_0}(U_0) = j_{U_1}(U_1)$. Moreover, $M_{U_0} = M_{U_1}$.**

Let $U$ be a normal ultrafilter on a measurable cardinal $\kappa$ and let $j : V\to M$ denote its ultrapower. Assume $2^\kappa = \kappa^+$. Let $\mathbb P$ be the Easton product $\prod_{\delta\in I}\text{Add}(\delta,1)$ where $I\subseteq\kappa$ is a $U$-null set of cardinals.
Let $\mathbb Q = j(\mathbb P)$ and let $\mathbb Q/\mathbb P$ denote the product  $\prod_{\delta\in j(I)\setminus \kappa}\text{Add}(\delta,1)$ as computed in $M$. Thus $\mathbb Q \cong \mathbb P\times (\mathbb Q/\mathbb P)$. Since $\kappa\notin j(I)$, $\mathbb Q/\mathbb P$ is ${\leq}\kappa$-closed and $j(\kappa)$-cc in $M$, so by standard arguments, one can construct an $M$-generic filter $G\subseteq \mathbb Q/\mathbb P$ in $V$.

Let $H\subseteq \mathbb P$ be a $V$-generic filter. Let $j_0:V[H]\to M[H\times G]$ be the unique lift of $j$ such that $j_0(H) = H\times G$. Let $\sigma_{\alpha,\kappa}$ denote the automorphism of $\mathbb P$ given by $$\sigma_{\alpha,\kappa}((p_\delta)_{\delta\in I}) = (p_\delta)_{\delta\in I\cap \alpha}{}^\frown(p^*_\delta)_{\delta\in I\setminus \alpha}$$
where for $q\in \text{Add}(\delta,1)$, $q^*$ denotes the result of flipping the bits in $q$. Denote the similar automorphism of $\mathbb Q$ by $\sigma_{\alpha,j(\kappa)}$. Let $j_1 : V[H]\to M[H\times G]$ be the lift of $j$ such that $j_1(H) = \sigma_{\kappa,j(\kappa)}(H\times G)$.

Now it's time to show $j_0(j_0) = j_1(j_1)$. Since $j_0(j_0)\restriction M = j_1(j_1)\restriction M$, it suffices to show that $j_0(j_0)(H\times G) = j_1(j_1)(H\times G).$ This follows from a long fun computation:

\begin{align*} 
   j_0(j_0)(H\times G) 
        &= j_0(j_0)(j_0(H))\\
        &= j_0(j_0(H))\\
        &= j_0(H\times G)\\
        &= j_0(H)\times j(G)\\  
        &= H\times G\times j(G)\\
        &= \sigma_{\kappa,j(j(\kappa))} \circ \sigma_{\kappa,j(j(\kappa))}(H\times G\times j(G))\\
        &= \sigma_{\kappa,j(j(\kappa))}\circ \sigma_{j(\kappa),j(j(\kappa))} (\sigma_{\kappa,j(\kappa)}(H\times G)\times j(G))\\
        &= \sigma_{\kappa,j(j(\kappa))}\circ \sigma_{j(\kappa),j(j(\kappa))} (j_1(H\times G))\\
        &= \sigma_{\kappa,j(j(\kappa))} (j_1(\sigma_{\kappa,j(\kappa)}(H\times G)))\\
        &= \sigma_{\kappa,j(j(\kappa))}(j_1(j_1(H)))\\ 
        &= \sigma_{\kappa,j(j(\kappa))}(j_1(j_1)(j_1(H)))\\
        &= j_1(j_1)(\sigma_{\kappa,j(\kappa)}(j_1(H)))\\
        &= j_1(j_1)(j_0(H))\\
        &= j_1(j_1)(H\times G)
\end{align*}


Finally, let $U_0$ and $U_1$ be the normal ultrafilters derived from $j_0$ and $j_1$. Since $j_0(j_0) = j_1(j_1)$, $j_0(U_0) = j_1(U_1)$, as desired.

**Second I'll sketch a proof that under the Ultrapower Axiom, the answer to your question is yes for arbitrary countably complete ultrafilters.**

I need the following theorem: *Assume UA, and let $U_0$ and $U_1$ be countably complete ultrafilters with ultrapowers $j_0 :V\to M_0$ and $j_1 :V\to M_1$. Then $j_1(U_0)$ belongs to $M_0$ if and only if there is an internal ultrapower embedding $k : M_0\to M_1$ such that $k\circ j_0 = j_1$.* (The reverse direction is obvious since $j_1(U_0) = k(j_0(U_0)).$)

Now assume UA. Suppose $U_0$ and $U_1$ are ultrafilters, $j_0 :V\to M_0$ and $j_1:V\to M_1$ are their ultrapowers, and $j_0(U_0) = j_1(U_1)$. I'll show $U_0 = U_1$. Assume without loss of generality that $U_0$ and $U_1$ are uniform ultrafilters on cardinals. Under UA, the constructibility preorder on such ultrafilters is linear (in fact, a prewellorder), so breaking symmetry, assume that $U_0\in L(U_1)$. Then $j_1(U_0)\in L(j_1(U_1)) = L(j_0(U_0)) \subseteq M_0$. Therefore by the theorem, there is an internal ultrapower embedding $k : M_0\to M_1$ such that $k\circ j_0 = j_1$. In particular, this means $U_0\leq_\text{RK} U_1$. Recall that the Rudin-Keisler order is wellfounded and note: $$|j_0(U_1)|_\text{RK}^{M_0}\leq k(|j_0(U_1)|_\text{RK}^{M_0}) = |j_1(U_1)|_\text{RK}^{M_1} = |j_0(U_0)|_\text{RK}^{M_0}$$ By the elementarity of $j_0$, the Rudin-Keisler rank of $U_1$ is less than or equal to that of $U_0$. Since $U_0\leq_\text{RK} U_1$, we must have $U_0\equiv_\text{RK} U_1$. In other words, $j_0 = j_1$, as desired.