I am looking for the largest possible class of regular graphs such that it is possible to form a basis out of the eigenvectors of the graph's adjacency matrix with elements of each vector having the same absolute magnitude (they would differ only in complex phase).
A vector $v=(1,1, \ldots, 1)^{T}$ is an eigenvector of the adjacency matrix of a graph $G$ if and only if $G$ is regular. What about other eigenvectors? I know examples of regular graphs, e.g., finite Cartesian integer lattices hypercubes on a integer lattice with with periodic boundary conditions, such that there is a complete set of eigenvectors with equal absolute values of the vector components (e.g., basis vectors for discrete Fourier transform). I am looking for a more general set of regulars graphs with such property.
This isn't a full answer but some partial results on constructing non-isomorphic graphs that share a basis of eigenvectors with coordinates in $\{-1,1\}$ including an eigenvector of all 1's.
Given a simple undirected graph $G$ of order $n$ the adjacency matrix $A$ will be a symmetric (that is $A^T=A$) $n \times n$ matrix. So the spectral theorem applies and A is orthogonally diagonizable. That is $A=PDP^T$ where $D$ is diagonal with real eigenvalues along the diagonal and the columns of $P$ form an orthonomal basis for $\mathbb{R}^n$. This means $P$ is an orthogonal matrix $P^TP=I$
Suppose $G$ meets the criteria of the problem then we can diagonalize $G=VDV^{-1}$ where $D$ is the same diagonal matrix as in the spectral theorem and $V$ is a matrix whose columns consist of a eigenvectors of $A$ that form a basis for $\mathbb{R}^n$ and have coordinates in $\{-1,1\}$ and includes the all ones eigenvector. The matrix $V$ is almost certainly not an orthogonal matrix unless there's some incredibly lucky graph.
Given a fixed basis of eigenvectors hence a fixed invertible matrix $V$ the only way we can have a non-isomorphic graph with the same eigenvectors is by having different eigenvalues corresponding to each vector. In particular the eigenvalue corresponding to the 1's eigenvector will be the regularity degree $d$ of the graph.
So given a graph $G$ with adjacency matrix $A=VDV^{-1}$ where the columns of $V$ form a basis for $\mathbb{R}^n$ of eigenvectors of $A$ whose coordinates are in $\{-1,1\}$ and first column is all 1's, i propose constructing other examples by changing the diagonal matrix $D$ until $VDV^{-1}$ is a matrix with coordinates in $\{0,1\}$ and zeros on the diagonal. I haven't found a method better than guessing and checking but I've produced a couple of examples this way. Here is one:
Take $G$ to be the cube graph.
This graph has adjacency matrix
$$A=
\begin{bmatrix}
0 & 1 & 1 & 0 & 1 & 0 & 0 & 0 \\
1 & 0 & 0 & 1 & 0 & 1 & 0 & 0 \\
1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 \\
0 & 1 & 1 & 0 & 0 & 0 & 0 & 1 \\
1 & 0 & 0 & 0 & 0 & 1 & 1 & 0 \\
0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 & 1 & 0 & 0 & 1 \\
0 & 0 & 0 & 1 & 0 & 1 & 1 & 0
\end{bmatrix}=VDV^{-1}$$
for
$
V=\left[\begin{array}{rrrrrrrr}
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & -1 & 1 & 1 & -1 & 1 & -1 & -1 \\
1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 \\
1 & 1 & 1 & -1 & -1 & -1 & -1 & 1 \\
1 & -1 & -1 & 1 & 1 & -1 & -1 & 1 \\
1 & 1 & -1 & 1 & -1 & -1 & 1 & -1 \\
1 & 1 & -1 & -1 & 1 & 1 & -1 & -1 \\
1 & -1 & -1 & -1 & -1 & 1 & 1 & 1
\end{array}\right]
$, and
$
D=\left[\begin{array}{rrrrrrrr}
3 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & -3 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & -1
\end{array}\right]
$
Setting
$D'=\left[\begin{array}{rrrrrrrr} 4 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & -4 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{array}\right]$
gives
$VD'V^{-1}= \left[\begin{array}{rrrrrrrr} 0 & 1 & 1 & 0 & 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & 1 & 0 & 1 & 1 & 0 \\ 1 & 0 & 0 & 1 & 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 & 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & 1 & 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 & 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 & 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & 1 & 0 & 1 & 1 & 0 \end{array}\right]$
Which has a corresponding $4$-regular graph that $G'$ that can't be isomorphic because the cube graph is $3$-regular. Here's a plot of said graph
I've successfully done this strategy on the 4-dimensional cube graph as well. But admittedly, I don't have any better ideas for a more general method.
Edit: I've improved my method slightly, but it won't scale well. Instead of guessing and checking I manually entered 5 as the degree of regularity and considered the matrix as a multivariate polynomial of the other eigenvalues. The condition of the diagonals have to be 0 and the other entries are in $\{0,1\}$ can be turned into a multivariable polynomial system of equations that I used Gröbner basis to solve. Up to isomorphism there is exactly one $5$-regular graph:
$ A=\left[\begin{array}{rrrrrrrr} 0 & 1 & 1 & 1 & 0 & 1 & 1 & 0 \\ 1 & 0 & 1 & 1 & 1 & 0 & 0 & 1 \\ 1 & 1 & 0 & 1 & 1 & 0 & 0 & 1 \\ 1 & 1 & 1 & 0 & 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 & 0 & 1 & 1 & 1 \\ 1 & 0 & 0 & 1 & 1 & 0 & 1 & 1 \\ 1 & 0 & 0 & 1 & 1 & 1 & 0 & 1 \\ 0 & 1 & 1 & 0 & 1 & 1 & 1 & 0 \end{array}\right] $ and $ D= \left[\begin{array}{rrrrrrrr} 5 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & -3 \end{array}\right] $
The GB approach also found a 2nd non-isomorphic 4-regular graph:
$$
\left(\begin{array}{rrrrrrrr}
0 & 1 & 0 & 1 & 0 & 1 & 1 & 0 \\
1 & 0 & 1 & 0 & 1 & 0 & 0 & 1 \\
0 & 1 & 0 & 1 & 1 & 0 & 0 & 1 \\
1 & 0 & 1 & 0 & 0 & 1 & 1 & 0 \\
0 & 1 & 1 & 0 & 0 & 1 & 0 & 1 \\
1 & 0 & 0 & 1 & 1 & 0 & 1 & 0 \\
1 & 0 & 0 & 1 & 0 & 1 & 0 & 1 \\
0 & 1 & 1 & 0 & 1 & 0 & 1 & 0
\end{array}\right)
\left(\begin{array}{rrrrrrrr}
4 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 2 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & -2 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & -2 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & -2
\end{array}\right)
$$