Lemma 1: Let $(X, \|\cdot\|)$ be an infinite-dimensional normed space. Then there exists a linear bijection $T$ from $X$ onto itself which is not a bounded linear operator on $(X, \|\cdot\|)$.
Proof: Let $\phi$ be an unbounded linear functional on $(X, \|\cdot\| )$. For a proof that such linear functionals exist, see this post. Choose $y\in X$ such that $\phi (y) = 1$. (Note that $y\neq 0$.) Define $T:X\to X$ by $T(x) := x - 2\phi (x)y$. Observe that $T$ is linear because $\phi$ is linear. If $x\in X$, then
\begin{align*}
T(T(x)) &= T(x) - 2\phi (T(x))y \\
&= (x - 2\phi (x)y) - 2\phi (x - 2\phi (x)y)y \\
&= x - 2\phi (x)y - 2[\phi (x) - 2\phi (x)\phi (y)]y \\
&= x - 2\phi (x)y - 2[-\phi (x)]y \\
&= x - 2\phi (x)y + 2\phi (x)y \\
&= x.
\end{align*}
Therefore $T$ is invertible with $T^{-1} = T$.
Suppose for a contradiction that $T$ is a bounded linear operator on $(X, \|\cdot\|)$. Then there is some $C>0$ such that $\|T(x)\| \leq C\|x\|$ for all $x\in X$. It follows for any $x\in X$ that
\begin{align*}
|\phi (x)| &= \|2y\|^{-1} |\phi (x)| \|2y\| \\
&= \|2y\|^{-1} \|2\phi (x) y\| \\
&= \|2y\|^{-1} \|(2\phi (x) y - x) + x\| \\
&\leq \|2y\|^{-1} (\|T(x)\| + \|x\|) \\
&\leq \|2y\|^{-1} (C\|x\| + \|x\|) \\
&= [\|2y\|^{-1}(C+1)] \|x\|.
\end{align*}
This shows that $\phi$ is a bounded linear functional on $(X, \|\cdot\|)$ which is a contradiction. Therefore $T$ cannot be a bounded linear operator on $(X, \|\cdot\|)$.
Lemma 2: Let $(Y, \|\cdot\|_{Y} )$ be a normed space. Suppose $X$ is another vector space over the same field as $Y$ and $T:X\to Y$ is an injective linear map. Define $\|\cdot \|_{X}:X\to [0, \infty )$ by $\|x\|_{X} := \|T(x)\|_{Y}$. Then $\|\cdot \|_{X}$ is a norm on $X$. If $\|\cdot\|_{Y}$ is a complete norm and $T$ is also surjective, then $\|\cdot\|_{X}$ is also a complete norm.
Proof: Let $x\in X$ such that $\|x\|_{X} = 0$. Then $\|T(x)\|_{Y} = 0$ and $T(x) = 0$ since $\|\cdot\|_{Y}$ is a norm on $Y$. As $T$ is injective, it follows that $x=0$.
Next, if $x,y\in X$ and $\alpha \in \mathbb{K}$, then by the linearity of $T$,
\begin{equation}
\|x+y\|_{X} = \|T(x+y)\|_{Y} = \|T(x) + T(y)\|_{Y} \leq \|T(x)\|_{Y} + \|T(y)\|_{Y} = \|x\|_{X} + \|y\|_{X}
\end{equation}
and
\begin{equation}
\|\alpha x \|_{X} = \|T(\alpha x) \|_{Y} = \| \alpha T(x) \|_{Y} = |\alpha | \|T(x)\|_{Y} = |\alpha | \|x\|_{X}
\end{equation}
hold. It follows that $\|\cdot\|_{X}$ is a norm on $X$.
Now suppose $\|\cdot\|_{Y}$ is a complete norm. Let $(x_{n})_{n\in\mathbb{N}}$ be a Cauchy sequence in $(X, \|\cdot\|_{X})$. Then
\begin{align*}
\|x_{m} - x_{n}\|_{X} = \|T(x_{m} - x_{n}) \|_{Y} = \|T(x_{m}) - T(x_{n})\|_{Y}
\end{align*}
for all $m,n\in\mathbb{N}$. This shows that $(T(x_{n}))_{n\in\mathbb{N}}$ is a Cauchy sequence in $(Y, \|\cdot\|_{Y})$. Since $\|\cdot\|_{Y}$ is a complete norm, the Cauchy sequence $(T(x_{n}))_{n\in\mathbb{N}}$ has a limit $y$ in $Y$. Choose $x\in X$ such that $T(x) = y$, which can be done because $T$ is surjective. Then
\begin{align*}
\|x_{n} - x\|_{X} = \|T(x_{n} - x)\|_{Y} = \|T(x_{n}) - T(x)\|_{Y} = \|T(x_{n}) - y\|_{Y}
\end{align*}
for each $n\in\mathbb{N}$. It follows that
\begin{equation}
\lim_{n\to\infty} \|x_{n} - x\|_{X} = \lim_{n\to\infty} \|T(x_{n}) - y\|_{Y} = 0
\end{equation}
and the sequence $(x_{n})_{n\in\mathbb{N}}$ converges to $x$ in $X$. Hence $\|\cdot\|_{X}$ is a complete norm on $X$.
Theorem: Let $X$ be an infinite-dimensional vector space with a complete norm $\|\cdot\|_{1}$. Then there exists another complete norm $\|\cdot\|_{2}$ on $X$ which is not equivalent to $\|\cdot\|_{1}$.
Proof: By Lemma 1, there exists a linear bijection $T:X\to X$ which is not a bounded linear operator on $(X, \|\cdot\|_{1})$. Define $\|\cdot\|_{2}:X\to [0, \infty )$ by $\|x\|_{2} := \|T(x)\|_{1}$. By Lemma 2, $\|\cdot\|_{2}$ is a complete norm on $X$. If the norms were equivalent, there would exist some $C>0$ such that $\|x\|_{2} \leq C \|x\|_{1}$ for all $x\in X$. But this would imply that $\|T(x)\|_{1} \leq C\|x\|_{1}$ for all $x\in X$ which cannot occur since $T$ is not a bounded linear operator on $(X, \|\cdot\|_{1})$. Therefore the norms $\|\cdot\|_{1}$ and $\|\cdot\|_{2}$ are not equivalent.
