**Prolog.**
As the arguments below are somewhat technical and probably not too interesting for many readers, I would like to point out that such problems can be quite subtle. The crucial problem (which could also be interesting for Banach spacers) is that the unique extension of a linear *injection* between normed spaces to the completions need not be injective! (Take a discontinuous linear functional $f$ on a Banach space $(X,\|\cdot\|)$ and consider identity from $X$ endowed with the stronger norm $\|x\|+|f(x)|$ to $(X,\|\cdot\|)$.)

A consequence of this is that even if a Fréchet space has continuous norms it is in general not possible to write it as an intersection of Banach spaces (i.e., it need not be a *countably normed* space -- which is a rather unfortunate terminology).
It was an observation of Pelczynski that an example of a nuclear Fréchet space with a continuous norm which is not countably normed would solve a problem of Grothendieck about the bounded approximation property for Fréchet spaces. Such examples where found in the eighties by Bellenot, Dubinsky, and Vogt.

**The answer.** Yes, this is possible. The main point in the argument is that any two *reduced* representations factorize through each other where a reduced representaion $F=\lim\limits_\leftarrow (F_i,f_j^i)$ is given by an increasing sequence of seminorms $p_i$ defining the topology such that $F_i$ is the Hausdorff completion of $(F,p_i)$ (first factor out the kernel of $p_i$ then take the completion) and $f_j^i:F_{j}\to F_i$ is the canonical map (associated to the identity $(F,p_j)\to (F,p_i)$) for $j\ge i$. Given two reduced representaions $F=\lim\limits_\leftarrow (F_i,f_j^i)$ and $F=\lim\limits_\leftarrow (H_i,h_j^i)$ there are $j(i)>i$ such that
$f_{j(i)}^i$ factorizes through some $H_k$ and $h_{j(i)}^i$ factorizes through some $F_k$. Passing to a subsequence we can thus assume that we have factorizations
$$
\cdots \leftarrow F_{i-1} \stackrel{\phi_{i-1}}{\leftarrow} H_i
\stackrel{\eta_{i}}{\leftarrow} F_i \leftarrow \cdots
$$
with $f_i^{i-1}=\phi_{i-1}\circ \eta_i$ and $h_{i+1}^i=\eta_i\circ \phi_i$.

Assume now that the $f_j^i$ are injective (because $F$ is countably normed) which implies that $\eta_i$ are also injective and that $h_j^i$ are trace class between Hilbert spaces (because $F$ is nuclear). Let $L_{i+1}$ be the kernel of $h_{i+1}^{i}$. The injectivity of $\eta_i$ implies that $L_{i+1}$ is the kernel of $\phi_i$ and hence the canonical map $\overline{h}_{i+1}^i:H_{i+1}/L_{i+1} \to H_i$ factorizes over $F_i$. Denoting the quotient map $q_{i+1} : H_{i+1}\to H_{i+1}/L_{i+1}$ we get the sequence
$$
\cdots\leftarrow F_{i-1} \leftarrow H_i/L_i \stackrel{q_i}{\leftarrow} H_i \stackrel{\eta_i}{\leftarrow} F_i \stackrel{\overline{h}_{i+1}^i}{\leftarrow} H_{i+1}/L_{i+1} \leftarrow \cdots
$$
One can then check that the map $H_{i+1}/L_{i+1}\to H_i/L_i$ is injective. Finally this implies that $H_{i+2}/L_{i+2}\to H_i/L_i$ is injective and trace class because it factorizes over a trace class operator.