First of all convince yourself that base change of a symmetric bilinear form corresponds to replacing the representing matrix $S$ by the matrix $P^TSP$, where $P \in GL_{d}(k)$ is the base change matrix between the two bases. The matrices $S$ and $P^TSP$ are called congruent to each other. Check that congruence defines an equivalence relation.
Over any field in which $2 \neq 0$, any symmetric matrix is congruent to a diagonal matrix. See Will Jagy's exposition here, with many examples and links with more examples.
For example, check that your matrix $\pmatrix{0&I_n\\I_n&0}$ is congruent to $\pmatrix{2I_n&0\\0&-\frac12 I_n}$ via $P = \pmatrix{I_n&-\frac12I_n\\I_n&\frac12I_n}$.
Now a diagonal matrix $\mathrm{diag}(d_1, ..., d_r)$ is congruent to $\mathrm{diag}(c_1^2d_1, ..., c_r^2d_r)$ via $P:=\mathrm{diag}(c_1, ..., c_r)$. In other words, we can change each of the diagonal entries by an appropriate square.
For example, over the field $\mathbb R$, the matrix $\pmatrix{2I_n&0\\0&-\frac12 I_n}$ is further congruent to $\pmatrix{I_n&0\\0& \color{red}{-} I_n}$ (and hence so is the original matrix $\pmatrix{0&I_n\\I_n&0}$: in the language of Sylvester's theorem, the form we are looking at here when considered over a real vector space has signature $(n,n)$).
But of course over an algebraically closed field $k$ (or any field in which all elements are squares), we can just make all diagonal entries $=1$ or $=0$. And zeros occur if and only if the form is degenerate. In particular, up to base change there is only one non-degenerate symmetric bilinear forms over an algebraically closed field $k$ of characteristic $\neq 2$. In particular, the one you have there is it, since it is congruent to $\pmatrix{I_n&0\\0&I_n}$ (note it is congruent to the identity matrix over all fields in which $2$ and $-\frac12$ are squares).
