Do equal eigenvalues in a SS matrix imply algebraic equivalence? In state space (SS) theory, the concept algebraic equivalence means, for matrix $A\in\mathbb{R}^{n\times n}$, column vector $B\in \mathbb{R}^n$ and row vector $C\in \mathbb{R}^n$, there is an invertible matrix $T$ such that
\begin{align}
\bar{A}&=TAT^{-1}\\
\bar{B}&=TB \tag{1}\label{eqtrue}\\
\bar{C}&=CT^{-1}
\end{align}
A method of solving this problem is provided as follows:
Solve $\begin{bmatrix}\bar{B} & \bar{A}\bar{B} & ... & \bar{A}^{n-1}\bar{B}\end{bmatrix} = T \begin{bmatrix}B & AB & ... & A^{n-1}B\end{bmatrix}$ for $T$ (under the assumption that $\begin{bmatrix}B & ... &A^{n-1}\end{bmatrix}$ and $\begin{bmatrix}\bar{B} & ...\end{bmatrix}$ are both full rank). Verify above relations using the $T$ obtained, that is, one system can be transformed in the other using $T$.

However, seeing as the next chapter discussed similarity transforms to a Jordan form, I had an itching feeling that maybe this could also be done using eigenvalues. I conjectured that maybe the following also implies algebraic equivalence:
\begin{equation}
\mathrm{eig}\left(\left[\begin{array}{c|c}
A & B\\\hline C & 0
\end{array} \right]\right) = \mathrm{eig}\left(\left[\begin{array}{c|c}
\bar{A} & \bar{B}\\\hline \bar{C} & 0
\end{array} \right]\right)\tag{2}\label{eq1}
\end{equation}
I.e., the set of eigenvalues of a matrix composed of writing $A,B,C$  in a square matrix (let's call it $SS$) equals the set of eigenvalues of the matrix composed of $\bar{A},\bar{B},\bar{C}$ (let's call that one $\bar{SS}$). For the examples I could find, this held up, but I fear that this may just be due to how these examples are 'cooked up'; and indeed, if this method would work, I should expect that we would be taught this method. I would love to find out how I could (dis)prove this, but I'm not sure how. I'm thinking along the lines of:
Construct a matrix $T^*$ as follows: 
\begin{align}T^*&=\left[\begin{array}{c|c}
T & 0\\\hline 0 & 1
\end{array} \right]& \to (T^*)^{-1}&=\left[\begin{array}{c|c}
T^{-1} & 0\\\hline 0 & 1
\end{array} \right]\end{align}
Then, we get
\begin{equation}
T^*\left[\begin{array}{c|c}
A & B\\\hline C & 0
\end{array} \right](T^*)^{-1} = \left[\begin{array}{c|c}
TAT^{-1} & TB\\\hline CT^{-1} & 0
\end{array} \right] \overset{?}= \left[\begin{array}{c|c}
\bar{A} & \bar{B}\\\hline \bar{C} & 0
\end{array} \right]
\end{equation}
Now, we need to see if there is a $T^*$ such that the above holds. From here on I'm speculating: I figured that there would be such a $T^*$ iff both matrices $SS$ and $\bar{SS}$ could be brought into the same diagonal form, so in other words, if \eqref{eq1} holds. 
Furthermore, the transformation matrix should have the same structure as $T^*$. So, let $M$ be the matrix bringing $SS$ into diagonal form, and $\bar{M}$ the matrix bringing $\bar{SS}$ into diagonal form, $M\bar{M}^{-1}$ should have zeroes on the last column and row and a final $1$ on the diagonal. But, here I'm stuck; how could I prove such a thing? Or do I even need to prove the same structure, or is condition \eqref{eq1} already sufficient, and will the desired structure follow naturally?
How can I (dis)prove that the above relation \eqref{eq1} implies algebraic equivalence as in \eqref{eqtrue}?

To address your main point: no, it is not enough for the eigenvalues to match.  In particular: the system with
$$
A = \pmatrix{0&1\\0&0}, \quad B^T = C = \pmatrix{0&0}
$$
produces a matrix whose eigenvalues are all $0$. However, we would get the same result with $A = 0$.

However, I have a feeling that it is indeed enough to check whether the two partitioned matrices $M = \left[ \begin{smallmatrix}
A&B\\C&0
\end{smallmatrix}\right]$ are similar.  Also, it's interesting to note that the lower-right entry could be anything (as long as we're consistent), such as $1$ instead of $0$.

Observation: if we take $T_0 = [B \quad AB \quad \cdots \quad A^{n-1}B]$, then we find that
$$
\pmatrix{T_0\\&1}^{-1} \pmatrix{A&B\\C&0} \pmatrix{T_0\\&1} = \pmatrix{\bar A & \bar B\\ \bar C & 0}
$$
where $\bar A$ is the transpose of that from the associated controllable canonical form, and $\bar B = (1,0,\dots,0)^T$.
