Consider a matrix $A \in \Bbb F^{n\times n}$ that is row equivalent to a matrix $B \in \Bbb F^{n\times n}$. This means that there exists an invertible matrix $Σ \in \Bbb F^{n\times n} : B=ΣΑ$
Is it true that these two matrices are also similar? Can we find an invertible $P \in \Bbb F^{n\times n}: B=P^{-1}AP$ ?
No, consider $$\begin{align} A &= \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix} \\ B &= \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \end{align}$$ which are row equivalent (you can get $B$ by subtracting $A$'s first row from its second row).
However, they are not similar since they have different eigenvalues (you can find a lot of properties for these matrices that are not the same, e.g. determinant, characteristic polynomial, etc, that should be the same for similar matrices). The eigenvalues of $A$ are $$\begin{align} \lambda_1 &= \frac{1}{2}\left(3 + \sqrt{5} \right) \\ \lambda_2 &= \frac{1}{2}\left(3 - \sqrt{5} \right) \end{align}$$ and $B$ only has one eigenvalue, namely $1$.
Counter-examples abound! Let $A \in \mathbb{F^{n \times n}} $ be any invertible matrix, $A \not= I$. Then $A^{-1}$ factors into elementary matrices which row reduce $A$ to the identity. But $A$ is not similar to the identity because the conjugacy class of $I$ contains only $I$ itself ($P^{-1}IP = I$ always).
In slightly more generality, we have the problem that not all invertible matrices are similar, but all invertible matrices are equivalent up to row operations. For any invertible matrices $A$ and $B$ we can rowreduce $B$ to $A$ with the matrix $AB^{-1}$. But note that conjugation preserves eigenvalues (similar matrices represent the same linear transformation) and we know that not all invertible matrices have the same eigenvalues.