Another Hadamard matrix of order 4?

Question

Wikipedia states that there is, up to equivalence, a unique Hadamard matrix of order 4, namely

$$ \def\p{\phantom+} \begin{pmatrix} \p1&\p1&\p1&\p1 \\ \p1&-1&\p1&-1 \\ \p1&\p1&-1&-1 \\ \p1&-1&-1&\p1 \end{pmatrix}.$$

As equialence operations are allowed negating some rows/columns or interchanging some rows/columns. But isn't the following another Hadamard matrix of order 4 that cannot be obtained in this way?

$$ \begin{pmatrix} -1&\p1&\p1&\p1 \\ \p1&-1&\p1&\p1 \\ \p1&\p1&-1&\p1 \\ \p1&\p1&\p1&-1 \\ \end{pmatrix}.$$

Answer 1

It can be obtained that way. Start at the first matrix. Negate first row to get: $$ \left( \begin{matrix} -1 & -1 & -1 & -1 \\ 1 & -1 & 1 & -1 \\ 1 & 1 & -1 & -1 \\ 1 & -1 & -1 & 1 \\ \end{matrix} \right) $$ Negate colums 2,3,4 to get: $$ \left( \begin{matrix} -1 & 1 & 1 & 1 \\ 1 & 1 & -1 & 1 \\ 1 & -1 & 1 & 1 \\ 1 & 1 & 1 & -1 \\ \end{matrix} \right) $$ Swap colums 2 and 3. And you get the second matrix.