Using matrix multiplication,
$$\begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1
\end{pmatrix}
=
\begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & 0 & i & 0 \\
0 & i & 0 & 0 \\
0 & 0 & 0 & 1
\end{pmatrix}
\begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & -i & 0 & 0 \\
0 & 0 & -i & 0 \\
0 & 0 & 0 & 1
\end{pmatrix}
$$
So the question is which gates does the second matrix in the matrix multiplication on the right represent?
A starting point is that the phase gate is $\begin{pmatrix} 1 & 0 \\ 0 & i \end{pmatrix}$ and that the Z-gate is $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$, and $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & i \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & -i \end{pmatrix}$. This gets us the upper left portion of the second matrix.
Once we have $\begin{pmatrix} 1 & 0 \\ 0 & -i \end{pmatrix}$, we can use the X-gate $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ and pre-and post-multiply it:$\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & -i \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} -i & 0 \\ 0 & 1 \end{pmatrix}$. This gets the lower right portion of the second matrix.