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.
