If two matrices have the same eigenvalues and eigenvectors are they equal?

The two matrices $\left[ \begin{array}{cc} 1 & 0 \\ 1 & 1 \end{array} \right] $ and $\left[ \begin{array}{cc} 1 & 0 \\ 200 & 1 \end{array} \right] $ both have the same eigenvalues and eigenvectors, but they are nowhere near equal to each other.

But if an $n\times n$ matrix has $n$ distinct eigenvalues or otherwise has a set of eigenvectors that form a basis of $\mathbb R^n,$ then the only matrix that has the same eigenpairs, i.e. the same eigenvectors, each with the same eigenvalue, is that same matrix. That is because a linear transformation is completely determined by what it does with a basis.

If, as the other answers assume, there are $n$ independent eigenvectors, i.e. if the matrices are diagonalizable, then the answer to your question is yes; using the $n$ independent eigenvectors as the base we see that the matrices (in that base) are identical and therefore they are identical in all bases. However, if the matrices are not diagonalizable, i.e. if there are no $n$ independent eigenvectors, then the matrices are not necessarily the same. For example:

$$A = \begin{pmatrix} 1&1\\ 0&1\\ \end{pmatrix}\;\;\text{and}\;\; B = \begin{pmatrix} 1&2\\ 0&1\\ \end{pmatrix}\,, $$

are different matrices that have the same eigenvectors with the same eigenvalues ($1$ is the only eigenvalue and its eigenspace is one-dimensional).

Besides needing a complete base of eigenvectors (as others pointed out), order is important if your "equal" is componentwise-equal. The following two matrices have the same set of eigenvalues and the same set of eigenvectors:

$$A = \begin{pmatrix} 2&0\\ 0&1\\ \end{pmatrix}\;\;\text{and}\;\; B = \begin{pmatrix} 1&0\\ 0&2\\ \end{pmatrix}\,. $$

(Conversely, having different eigenvectors does not necessarily mean the matrices are different. Any base of each eigenspace works as eigenvectors. (Think scaling or repeated eigenvalues.))