What is the Lie group that leaves this matrix invariant? What is the group that leaves 
\begin{equation}
Y = \bigoplus_{j=1}^{k} \alpha_j \mathbb{I}_{2n_j\times 2n_j}
\end{equation}
 invariant under congruence ($Y = XYX^T$) where $\alpha_j \in \mathbb{R}$ and $\alpha_j \neq 0$?
I have toyed with different variations of the special orthogonal group; my misunderstandings of how matrix groups relate to representations are fleshed out here.
 A: Answer from Matt Biesecker in comments. Sylvester's Law taken from wiki.
Sylvester's Law of Inertia:
Given some non-singular real symmetric matrix $A$, the group of matrices that satisfy
\begin{equation}
 XAX^T = A.
\end{equation}
is $\operatorname{O}(p,q)$ where $p$ and $q$ are the number of postive and negative eigenvalues of $A$ respectively.
Proof:
Real symmetric matrices can be diagonalised by congruence by an orthogonal matrix, so we know that there exists $Q$ such that
\begin{equation}
A = QEQ^T
\end{equation}
where $E$ is the diagonal matrix of eigenvalues. Given $W$ such that $W_{ii} = (\sqrt{|E_{ii}|})^{-1}$ we obtain 
\begin{equation}
E = WDW^T
\end{equation}
where $D$ is a diagonal matrix of $1$s and $-1$s. Thus
\begin{equation}
 XAX^T = (XQW)D(XQW)^T = (QW)D(QW)^T.
\end{equation}
and so
\begin{equation}
 \left((QW)^{-1}X(QW)\right)D\left((QW)^{-1}X(QW)\right)^T = D.
\end{equation}
The group that preserves $D$ under congruence is $\operatorname{O}(p,q)$ by definition. Thus the group of $X$ matrices is isomorphic to $\operatorname{O}(p,q)$.
Relation to question:
$Y$ is a nonsingular real symmetrix matrix.
