This is only a partial result showing that any counterexample $e$ must have $4 \mid e$ and have at least one odd prime factor, i.e. the statement holds for all $e=2t$ with $t\geq 3$ odd and $e=2^k$ for $k \geq 2$.
Proof. Assume $e=2t$ with $t$ odd and consider solutions to $\lambda(m)=2t$. Write prime factorization $m=2^{e_0}p_1^{e_1}\cdots p_k^{e_k}$ for some odd primes $p_i$. Then
$$
p_i-1\mid\varphi(p_i^{e_i})=\lambda(p_i^{e_i}) \mid \operatorname{lcm}(\lambda(2^{e_0}),\lambda(p_1^{e_1}),\dots,\lambda(p_k^{e_k}))=\lambda(m)=2t
$$
and thus $p_i=d+1$ for some $d \mid 2t$. So each solution consists of some subset of primes $$T=\{p_1,p_2,\dots,p_k\} \subseteq \{d+1;d \mid 2t, d+1 \text{ is odd prime}\}.$$ From now on we assume fixed set $T$ and inspect all possible values of exponents $e_i$.
We have \begin{align}\lambda(m)&=\operatorname{lcm}(\lambda(2^{e_0}),\lambda(p_1^{e_1}),\dots,\lambda(p_k^{e_k}))\\
&=\operatorname{lcm}(\lambda(2^{e_0}),p_1^{e_1-1}(p_1-1),\dots,p_k^{e_k-1}(p_k-1))\\
&=\operatorname{lcm}(\lambda(2^{e_0}),p_1^{e_1-1},\dots,p_k^{e_k-1},(p_1-1),\dots,(p_k-1))\\
&=\operatorname{lcm}(\lambda(2^{e_0}),p_1^{e_1-1}\cdots p_k^{e_k-1},\operatorname{lcm}(p_1-1,\dots,p_k-1))
\end{align}
So let $g=\operatorname{lcm}(p_1-1,\dots,p_k-1)=2^{u_0}p_1^{u_1}\cdots p_k^{u_k}\cdot v$ for some $0 \leq u_i$ and integer $v$ such that $(2p_1\cdots p_k,v)=1$. Hence
\begin{align}
\lambda(m)&=\operatorname{lcm}(\lambda(2^{e_0}),2^{u_0})\cdot \operatorname{lcm}(p_1^{e_1-1},p_1^{u_1})\cdots \operatorname{lcm}(p_k^{e_k-1},p_k^{u_k})\cdot v\\
&=\operatorname{lcm}(\lambda(2^{e_0}),2^{u_0})\cdot p_1^{\max(e_1-1,u_1)}\cdots p_k^{\max(e_k-1,u_k)}\cdot v\\
&=2t
\end{align}
If $2t$ cannot be written as $2\cdot p_1^{r_1}\cdots p_k^{r_k}\cdot v$, then for given $T$ we have zero solutions contributing to the total count. If on the other hand we can write $2t$ in the above form, comparing the exponents we get conditions for $e_i$. Also these conditions are independent, i.e. we can count the solutions for each $e_i$ separately and multiply them together, say $|A_{T,0}|\cdot|A_{T,1}|\cdots |A_{T,k}|$, where $A_{T,i}$ are all possible values for $e_i$ (for given set $T$). The total number of solutions is just sum over all $T$. We will now show that under our conditions $|A_{T,0}|=4$, hence by the above number of solutions is multiple of $4$.
Since $A_{T,0}$ is the set of integers $e_0$ such that $2=\operatorname{lcm}(\lambda(2^{e_0}),2^{u_0})$ (by comparing the exponents), we have $2^{u_0}\mid 2$ and so $u_0=0$ or $u_0=1$. However $u_0=0$ would imply $g=1$ and $m=2^{e_0}$. But then $\lambda(m)=2t$ must be power of $2$, impossible for $t \geq 3$ odd. So we must have $u_0=1$. Thus $\lambda(2^{e_0}) \mid 2$ and simple case work shows this is equivalent to $e_0 \in \{0,1,2,3\}$. Hence $A_{T,0}=\{0,1,2,3\}$ and $|A_{T,0}|=4$ as desired.
$$\tag*{$\blacksquare$}$$
As an example consider $\lambda(m)=6$. It has sixteen solutions $$m\in \{7, 9, 14, 18, 21, 28, 36, 42, 56, 63, 72, 84, 126, 168, 252, 504\}.$$ If we group them by odd prime factors, we have the following three groups:
$T=\{3\}:9,18,36,72$. All the values are exactly products of elements of $\{2^0,2^1,2^2,2^3\}$ and $\{3^2\}$.
$T=\{7\}:7,14,28,56$. All the values are exactly products of elements of $\{2^0,2^1,2^2,2^3\}$ and $\{7\}$.
$T=\{3,7\}:21,42,63,84,126,168,252,504$. All the values are exactly products of elements of $\{2^0,2^1,2^2,2^3\}$, $\{3,3^2\}$ and $\{7\}$.
Such grouping of solutions into sets for given $T$ works in general, unfortunately $|A_{T,0}|=4$ does not for $4\mid e$, so another idea/grouping is needed.
Added: Additional result for $e=2^k$, it can be shown by the methods above that
$$
n(2^k)=\begin{cases}
6 & k=1 \\
2^c\cdot(k+4) & 2^k+1 \text{ is prime } \\
2^c & \text{otherwise}
\end{cases}
$$
where $c$ is the number of primes among $2^1+1,2^2+1,\dots,2^{k-1}+1$ (Fermat primes). This also gives $4\mid n(2^k)$ for $k\geq 2$.
