The only eigenvalue of $A \in {M_n}$ is $\lambda = 1$. Why is $A$ similar to $A^k$? Suppose  that the only eigenvalue of $A \in {M_n}$ is $\lambda  = 1$. Why is $A$ similar to $A^k$ for each $k=1,2,3,\dots$?
 A: This is not true in general, it depends on the underlying field. Take $A=\pmatrix{ 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & - 1 & 0 }$. It has only one real eigenvalue (which is one), but is not similar to e.g. $A^4 = I_3$.

In the case of a complex field, we have to show that the Jordan forms of $A$ and $A^k$ are equal. From the construction of the Jordan form it follows that they are equal iff
$$
rank((\lambda I - A)^n) = rank((\lambda I - A^k)^n)
$$
for all eigenvalues $\lambda$ and all powers $n$. Since $\lambda=1$ is the only (complex) eigenvalue it remains to check this case.
Now it holds
$$
I-A^k = (I-A)(I + A + \dots A^{k-1}).
$$
Since $1$ is the only eigenvalue of $A$, $k$ is the only eigenvalue of $I + A + \dots A^{k-1}$, hence this matrix is invertible.
Multiplying with invertible matrices leaves rank invariant, hence
$$
rank( (I-A^k)^n ) =  rank( (I-A)^n(I + A + \dots A^{k-1})^n) = rank( (I-A)^n,
$$
which implies that the Jordan forms of $A$ and $A^k$ are equal.

I wonder, whether there is a more direct solution to this question, that is to construct invertible $T$ such that $A=TA^kT^{-1}$.
