Does the set of $n\times n$ invertible matrices with real coeffs where the sum of each row and column sums to one forms a group under multiplication?
Consider such set $S$, four things must hold for it to form a group:
Closure: $a\cdot b\in G,\ \forall a,b\in S$
Associativity: $a\cdot(b\cdot c)=(a\cdot b)\cdot c,\ \forall a,b,c\in S$
Identity: $\exists e\in S,\forall a\in S:e\cdot a = a\cdot e = a$
Inverse: $\forall a\in S,\exists a^{-1}\in S:(a^{-1})\cdot a = a\cdot (a^{-1}) = e$
The set $S$ with multiplication operation clearly is associative (it is a subset of all $n\times n$ matrices), has identity $I:S\to S$ and every element has an inverse (because the matrices in $S$ are invertible), but is it closed under the operation? The result of multiplying two invertible matrices whose rows and columns sums to one gives another invertible matrix whose rows and columns sums to one?
If the case is that yes, it forms a group, then the result tells that the order of $S$ divides the order of $\text{GL}_n(\Bbb R)$ (because of Lagrange).
Note that $S\neq\emptyset$, since $\operatorname{Id}_n\in S$.
Let $v=(1,1,\ldots,1)$. Then the set $S$ can be seen as$$\left\{M\in GL_n(\mathbb{R})\,\middle|\,M.v=v\text{ and }M^t.v=v\right\}.$$It is clear from this definition, and from the fact that $(MN)^t=N^tM^t$, that $S$ is closed with respect to the matrix product. Furthermore, if $M\in S$, then $M.v=v\implies v=M^{-1}.v$ and\begin{align}M^t.v=v\implies&(M^t)^{-1}.v=v\\\iff&(M^{-i})^t.v=v.\end{align}So, $M\in S\implies M^{-1}\in S$. Since $GL_n(\mathbb{R})$ is a group, this proves that $S$ is indeed a group.
As you have already been told, asserting that $\#S\mid\#GL_n(\mathbb{R})$ makes no sense here.