In this question, the OP asks about finding the matrix exponential of the matrix $$M=\begin{bmatrix}
1 & 1 & 1\\
1 & 1 & 1\\
1 & 1 & 1
\end{bmatrix}.$$ It works out quite nicely because $M^2 = 3M$ so $M^n = 3^{n-1}M$. The reason this occurs is that the vector $$v = \begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix}$$ is an eigenvector for $M$ (with eigenvalue $3$). The same is true of the $n\times n$ matrix consisting of all ones and the corresponding vector. With this in mind, I ask the following:
Can we find a standard form for an $n\times n$ matrix with the property that each of its columns are eigenvectors? (Added later: This may be easier to answer if we allow columns to be zero vectors. Thanks Adam W for the comment.)
What about the case when we require all the eigenvectors to correspond to the same eigenvalue?
The matrices in the second question are precisely the ones for which the calculation of the matrix exponential would be analogous to that for $M$ as above.
Added later: In Hurkyl's answer, he/she shows that an invertible matrix satisfying 1 is diagonal. For the case $n=2$, it is fairly easy to see that any non-invertible matrix satisfies 1 (which is generalised by the situation in Seirios's answer).
However, for $n > 2$, not every non-invertible matrix satisfies this property (as one may expect). For example,
$$M = \begin{bmatrix}
1 & 0 & 0\\
0 & 0 & 0\\
1 & 0 & 1
\end{bmatrix}.$$
Best Answer
If each of the columns of $A$ are eigenvectors of $A$, then
$$ AA = AD $$
where $D$ is the diagonal matrix of the corresponding eigenvalues. So the matrix exponential is easy to compute:
$$ A^n = A D^{n-1}$$
As a partial answer to describing which matrices have this property, if $A$ is invertible, then we must actually have $A = D$. Also, every diagonal matrix has the stated property.