I assume what you mean is that your model consists of many prisms, with each vertex of the prisms having a displacement vector, but these vertices are not shared among prisms, so we need only look at one prism at a time. Furthermore, I assume that the prisms can only undergo rigid body motions, meaning that their initial and final states are related by a Euclidean transformation.
Overview of rigid transformations
Formally, let's assume a definite order of all the vertices, so that we can arrange the displacement vectors into a matrix, each column of which is one displacement vector of a particular vertex. Call it $V$. $V$ has 3 rows and (typically) 12 columns. Let's also assume you can do something similar for the prism vertex position vectors, forming a matrix $P$ (we won't be interested in $P$ itself, but it's useful for the discussion. The initial position of the prism is specified by $P$, and the final position is $Q=P+V$. The fact that you only allow rigid transformations means that $P+V=RP+\mathbf{t}\mathbf{1}^T$, where $\mathbf{1}$ is the vector of all 1's (in this case typically of length 12), $R$ is a rotation matrix with determinant 1, and $\mathbf{t}$ is a translation vector. So you can see that really, the entire set of displacement vectors is specified by just $R$ and $\mathbf{t}$, which is 12 real numbers, with the implicit assumption that you already know $P$ the prism position.
Your problem basically boils down to computing a rigid transformation given only 3 points. See, for example this paper.
Using only 3 vertices
Suppose you only consider the first three columns of $P$ and $V$ (call them $P_3$ and $V_3$, correspondingly $Q_3$), which is like only looking at 3 of the vertices. Then we have $ Q_3=RP_3+\mathbf{t}\mathbf{1}^T$ The translation is very annoying, so let's get rid of it first. The centroid of the vertices is unaffected by the rotation, so let's work relative to the centroids. Call the centroids $p_c=\frac{1}{3}P_3\mathbf{1}$ and $q_c=\frac{1}{3}Q_3\mathbf{1}$. Define new translated matrices $ Q_c = Q_3 - q_c\mathbf{1}^T $ $ P_c = P_3 - p_c\mathbf{1}^T $ We now have $ Q_c=RP_c$. You can't just solve for $R = Q_cP_c^{-1}$ because the matrices $Q_c$ and $P_c$ will only have rank 2. You need to compute the SVD of $P_cQ_c^T=U\Sigma V^T$, and set $R = VU^T$. To calculate any $i$-th displacement vector, $ v_i = R(p_i - p_c) + q_c $
More robust scheme
If your displacements are not entirely consistent, then the rotation you compute may not be accurate (it may not be an exactly orthogonal matrix). It is instead best to store the $R$ matrix and the displaced centroid $q_c$ (redefined below) rather than just $V_3$. You want to compute the "best" $R$ given all the displacements you have. Following the linked paper, similarly compute the $3\times n$ centroid-relative matrices $ Q_c = Q - q_c\mathbf{1}^T $ $ P_c = P - p_c\mathbf{1}^T $ for analogous definitions of the centroids $q_c$ and $p_c$. You once again have $ Q_c=RP_c$, but now $Q_c$ and $P_c$ are wider than they are tall. The best $R$ in the least squares sense is $R=VU^T$ where $V$ and $U$ are derived from the singular value decomposition of $P_cQ_c^T=U\Sigma V^T$. If your simulations are self-consistent, $\Sigma$ should be the $3\times 3$ identity matrix, but due to rounding errors, it may be slightly off.
Example
Suppose you have three points $(2,-1,4)$, $(-3,0,1)$, and $(5,1,-3)$, with respective displacements $(-0.75,-0.3,-0.58)$, $(0.65,-0.24,-0.72)$, $(0.07,-0.92,0.38)$. We would then form the following matrices: $ P_3 = \begin{bmatrix} 2&-3&5\\-1&0&1\\4&1&-3\end{bmatrix} $ $ Q_3 = \begin{bmatrix} -0.75& 0.65&0.07 \\ -0.3& -0.24 &-0.92\\-0.58&-0.72&0.38\end{bmatrix}$ The centroids are $p_c=(4/3,0,2/3)$ and $q_c = (1.323,-0.487,0.36)$. The recentered matrices are $ P_c = \begin{bmatrix} 2/3 & -13/3 & 11/3 \\ -1 &0& 1\\10/3 & 1/3 & -11/3 \end{bmatrix} $ $ Q_c = \begin{bmatrix} -0.073 & -3.673 & 3.747 \\ -0.813 &0.246& 0.566\\3.06 & -0.08 & -2.98 \end{bmatrix} $ Then compute the SVD of $P_cQ_c^T$, letting $\{U,\Sigma,V\}=svd(P_cQ_c^T)$ (yes, this $V$ is a different $V$ now). Finally, $R = VU^T$ results in $ R = \begin{bmatrix} 0.990 & 0.075 & -0.122 \\ -0.071 &0.997& 0.036\\0.125 & -0.027 & 0.991 \end{bmatrix} $ Note that $R$ is almost the identity matrix (diagonal elements are almost 1, off diagonal are small) since our displacements were small. Note also that $RP_c \approx Q_c$. It's not exactly equal because the displacements I chose are only approximately corresponding to a rigid transformation (I rounded the numbers, so this introduces some stretch and shear).
Now, if we had a new point $p_4=(3,3,3)$, it's displacement is $ v_4 = q_c + R(p_4-p_c) - p_4 = (-0.087,-0.531,-0.199)$