Your claim is correct. In fact $\sigma=1$ is a singular value with multiplicity at least $m-2$.
If $\mathbf{A}$ has the controller canonical form
$$\mathbf{A}=\left[\matrix{0 & 1 & 0 & \cdots & 0\\ 0 & 0 & 1 & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & 0 & \cdots & 1\\ -a_0 & -a_1 & -a_2 & \cdots & -a_{m-1}}\right]\qquad \qquad (1)$$
then the characteristic polynomial of $\mathbf{A}$ is
$$p_{\mathbf{A}}(\lambda)=\det(\lambda I-\mathbf{A})=\lambda^m+a_{m-1}\lambda^{m-1}+\cdots+a_0$$
It holds true that $a_0=(-1)^m\prod_{i=1}^m{\lambda_i(\mathbf{A})}$ and since $|\lambda_i(\mathbf{A})|<1$ for all $i=1,\cdots,m$ we directly deduce that $|a_0|<1$.
Then we can write
$$\mathbf{A^TA}=\left[\matrix{a_0^2 & a_0a_1 & a_0a_2 & \cdots & a_0a_{m-1}\\ a_0a_1 & 1+a_1^2 & a_1a_2 & \cdots & a_1a_{m-1}\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ a_0a_{m-2} & a_1a_{m-2} & a_2a_{m-2} & \cdots & a_{m-2}a_{m-1}\\a_0a_{m-1} & a_1a_{m-1} & a_2a_{m-1} & \cdots & 1+a_{m-1}^2}\right]$$
which can be decomposed as
$$\mathbf{A^TA}=\mathbf{N}+\mathbf{a}\mathbf{a}^T$$
with
$$\mathbf{N}=diag\{0,1,1,\cdots,1\}\:, \quad\mathbf{a}=\left[\matrix{a_0 & a_1 & \cdots & a_{m-1}}\right]^T$$
If $\lambda$ is an eigenvalue of $\mathbf{A^TA}$ and $\mathbf{v}=\left[\matrix{v_1 & v_2 & \cdots & v_{m}}\right]^T$ the respective right eigenvector then
$$\lambda\mathbf{v}=\mathbf{A^TAv}=\mathbf{Nv}+\mathbf{a}\mathbf{a}^T\mathbf{v}$$
or equivalently
$$\lambda v_1 =(\mathbf{a}^T\mathbf{v})a_0\qquad \qquad \qquad \qquad \qquad \qquad\quad (2)\\
(\lambda-1) v_i=(\mathbf{a}^T\mathbf{v})a_{i-1}\:,\quad (i=2,\cdots,m) \qquad (3)$$
Define now the reduced vectors
$$\mathbf{a_r}=\left[\matrix{a_1 & a_2 &\cdots & a_{m-1}}\right]^T\:,\quad \mathbf{v_r}=\left[\matrix{v_2 & v_3 &\cdots & v_{m}}\right]^T$$ and the vector $\mathbf{e_i}$ which denotes the $i$-th column of the identity matrix ($i=1,2,\cdots,m$).
Case 1: $\quad\mathbf{a_r}=0$
Then from (2), (3) we have
$$\lambda v_1=a_0^2v_1\\ (\lambda-1)v_i=0\:,\quad (i=2,\cdots,m)$$
which mean that there are exactly 2 eigenvalues:
$\lambda_1=1$ with multiplicity $\mu_1=m-1$ and independent
eigenvectors $\mathbf{e_2},\cdots, \mathbf{e_m}$ and a simple
eigenvalue $\lambda_2=a_0^2<1$ with eigenvector $\mathbf{e_1}$.
Thus for Case 1 the maximum singular value of $\mathbf{A}$ is 1.
Case 2: $\quad\mathbf{a_r}\neq 0$
Then $\dim \ker(\mathbf{a_r}^T)=m-2$ (there are exactly $m-2$ independent vectors normal to $\mathbf{a_r}$) and from (2), (3) $\mathbf{A^TA}$ has an eigenvalue
$\lambda_1=1$ with multiplicity $\mu_1=m-2$ and independent
eigenvectors of the form $\left[\matrix{0 & \mathbf{v_r}^T}\right]^T$ with $\mathbf{v_r}\in \ker(\mathbf{a_r}^T)$.
We will examine now the behavior of the other 2 eigenvalues of $\mathbf{A^TA}$.
Case 2.A:$\quad a_0=0$
From (2), (3) we have
$$\lambda v_1 =0\qquad \qquad \qquad \qquad \\
(\lambda-1) \mathbf{v_r}=(\mathbf{a_r}^T\mathbf{v_r})\mathbf{a_{r}} \qquad $$
and therefore the other 2 eigenvalues are
$\lambda_2=0$ with eigenvector $\mathbf{e_1}$ and
$\lambda_3=1+\mathbf{a_r^Ta_r}$ with eigenvector $\left[\matrix{0&
\frac{\mathbf{a_r^T}}{\|\mathbf{a_r}\|}}\right]^T$
Thus, for Case 2.A there is exactly one singular value of $\mathbf{A}$ larger than one namely $\sqrt{1+\mathbf{a_r^Ta_r}}$.
Case 2.B:$\quad a_0\neq 0$
From (1) we can easily calculate that
$\det(\mathbf{A})=(-1)^ma_0$ and therefore
$$\det(\mathbf{A^TA})=(\det(\mathbf{A}))^2=a_0^2$$
i.e. $0\neq \det(\mathbf{A^TA})<1$. Thus,
$$0\neq
\det(\mathbf{A^TA})=\lambda_1^{m-2}\lambda_2\lambda_3=\lambda_2\lambda_3=a_0^2<1$$
and therefore at most one of the eigenvalues $\lambda_2,\lambda_3$ can be larger that one.