To emphasize that linearity of the map $T$ is not necessary, let's proceed with a careful development that avoids such an assumption. At the end we can point out the minor "simplification" in notation and definition that linearity allows.
Consider $T:V\to V$, where $V$ is a normed vector space with two equivalent norms, say $||\cdot||_1$ and $||\cdot||_2$ such that for some positive constants $a,b$:
$$ a ||x||_1 \le ||x||_2 \le b ||x||_1 \text{ for all }x\in V $$
We say that $T$ is a contraction map on $V$ with respect to norm $||\cdot||_1$ iff there exists a constant $c\in [0,1)$ such that for all $x,y \in V$:
$$ ||T(x) - T(y)||_1 \le c ||x - y||_1 $$
Our challenge is to show there exists a positive integer $k$ such that $T^k$ is also a contraction with respect to the second norm $||\cdot||_2$. To begin with one shows, by induction, that for all $x,y \in V$:
$$ ||T^k(x) - T^k(y)||_1 \le c^k ||x - y||_1 $$
The basis step $k=1$ is already in hand (as above). The induction step does not require any linearity assumption, as the following shows:
$$ \begin{align*}
||T^{k+1}(x) - T^{k+1}(y)||_1 &= ||T^k(T(x)) - T^k(T(y))||_1 \\
&\le c^k ||T(x) - T(y)||_1 \\
&\le c^{k+1} ||x - y||_1 \end{align*} $$
Here we have first rewritten the operator $T^{k+1}$ as a composition, then applied the induction hypothesis to the result, and finally used the base estimate to obtain an extra factor of $c$.
Now we have the following chain of inequalities:
$$ \begin{align*}
||T^k(x) - T^k(y)||_2 &\le b ||T^k(x) - T^k(y)||_1 \\
&\le bc^k ||x - y||_1 \\
&\le \frac{bc^k}{a} ||x - y||_2 \end{align*} $$
The desired contractive property for $T^k$ will follow when $k$ is chosen large enough that, since $0 \le c \lt 1$, the factor $c^k$ is small enough that:
$$ 0 \le \frac{bc^k}{a} \lt 1 $$
In other words pick $k$ such that $c^k \lt a/b$.
Based on some feedback I got in chat, Readers may have difficulty envisioning an actual example, so let's give one.
In a finite dimensional normed vector space, it turns out that all norms are equivalent. The converse is also true.
For simplicity we will consider the two-dimensional vector space $\mathbb{R}^2$, and the linear transformation $T$ of matrix multiplication by:
$$ \begin{pmatrix}
0 & 0 \\ 1 & 0
\end{pmatrix} $$
Obviously $T^2 = 0$ is a contraction with respect to the Euclidean norm (or with respect to any norm), but since:
$$ T\begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \end{pmatrix} $$
$T$ itself is not a contraction in the Euclidean norm $||\cdot||_2$. However a custom norm can be created with respect to which $T$ is contraction, such as in particular:
$$ || (x,y)^T ||_1 = \sqrt{x^2 + (y^2)/2} $$
Thus $||T(x,y)^T||_1 = || (0,x)^T ||_1 = \sqrt{0.5 x^2} \lt 0.8 || (x,y)^T ||_1 $, so in this custom norm $T$ is a contraction.
