The combinatorial class of $k$-ary rooted unlabeled trees is
$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}
\mathcal{T} = \mathcal{Z} +
\mathcal{Z} \times \textsc{SEQ}_{=k}(\mathcal{T})$$
Which gives the functional equation
$$T(z) = z + z \times T(z)^k$$
so that
$$z = \frac{T(z)}{1+T(z)^k}$$
We then have
$$n \times T_n = [z^{n-1}] T'(z)$$
and from the Cauchy Coefficient Formula
$$[z^{n-1}] T'(z) =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^n} T'(z) \; dz.$$
Now put $T(z) = w$ so that $T'(z) \; dz = dw$
and we obtain
$$\frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{(1+w^k)^n}{w^n} \; dw.$$
This is
$$T_n =
[[n\equiv 1 \bmod k]]
\frac{1}{n}
[w^{n-1}] (1+w^k)^n
\\ =
[[n\equiv 1 \bmod k]]
\frac{1}{n}
[w^{k\times (n-1)/k}] (1+w^k)^n
\\ =
[[n\equiv 1 \bmod k]]
\frac{1}{n}
[w^{(n-1)/k}] (1+w)^n.$$
We thus have for the answer
$$\bbox[5px,border:2px solid #00A000]{
[[n\equiv 1 \bmod k]]
\frac{1}{n}
{n\choose (n-1)/k}.}$$
With this answer we observe however that we can get a better formula
by adhering to the convention that the size of the tree is the number
of internal nodes rather than the total number of nodes. This yields
the class
$$\mathcal{T} = \mathcal{E} +
\mathcal{Z} \times \textsc{SEQ}_{=k}(\mathcal{T})$$
or
$$T(z) = 1 + z \times T(z)^k$$
so that
$$z = \frac{T(z)-1}{T(z)^k}$$
We get
$$\frac{1}{2\pi i}
\int_{|w-1|=\gamma}
\frac{w^{kn}}{(w-1)^n} \; dw
= \frac{1}{2\pi i}
\int_{|w-1|=\gamma}
\frac{1}{(w-1)^n} \sum_{q=0}^{kn} {kn\choose q} (w-1)^q\; dw.$$
We thus have for the answer in terms of the number of internal nodes
$$\bbox[5px,border:2px solid #00A000]{
\frac{1}{n}
{kn\choose n-1}.}$$
We then have for $k=3$ (ternary trees) the sequence
$$1, 3, 12, 55, 273, 1428, 7752, 43263, 246675, 1430715, \ldots$$
which points to OEIS A001764, where we find
confirmation. Similarly, $k=4$ (quartic trees) gives
$$1, 4, 22, 140, 969, 7084, 53820, 420732, 3362260, 27343888, \ldots$$
which points to OEIS A002293.
We may re-write the binomial coefficient as follows:
$$\frac{(kn)!}{n! \times ((k-1)n+1)!}$$
or
$$\bbox[5px,border:2px solid #00A000]{
\frac{1}{(k-1)n+1}
{kn\choose n}.}$$
This form is sometimes preferred because it is defined when $n=0$,
where it yields the value one (one empty tree). It also happens to
match the standard Catalan number formula when $k=2.$
Addendum. I did not read the question properly and answered
for rooted ordered unlabeled k-ary plane trees instead of rooted
labeled k-ary trees. The combinatorial class now becomes
$$\mathcal{T} = \mathcal{Z} +
\mathcal{Z} \times \textsc{SET}_{=k}(\mathcal{T})$$
or
$$T(z) = z + z \times \frac{1}{k!} T(z)^k$$
so that
$$z = \frac{T(z)}{1+T(z)^k/k!}.$$
The computation is the same as before and we obtain
(multiply by $n!$ for an EGF)
$$\bbox[5px,border:2px solid #00A000]{
[[n\equiv 1 \bmod k]]
\frac{(n-1)!}{k!^{(n-1)/k}}
{n\choose (n-1)/k}.}$$
We get for binary trees on $n=2m+1$ nodes
$$1, 3, 60, 3150, 317520, 52390800, 12843230400, 4382752374000,
\\ 1986847742880000, 1155153277710432000, 838011196011749760000,
\ldots $$
which is OEIS A036770. For ternary trees on
$n=3m+1$ nodes we find
$$1, 4, 420, 201600, 264264000, 734557824000, 3723191087616000,
\\ 31125877492469760000, 399532678960326912000000, \ldots $$
which is OEIS A036771. Quartic trees on
$n=4m+1$ nodes then yield
$$1, 5, 2520, 9909900, 150089940000, 6217438242015000,
\\ 574985352122181000000, 103753754577643425255000000, \ldots$$
which is OEIS A036772.
For convenience we may write $n=mk+1$ to do without the Iverson
bracket and get
$$\bbox[5px,border:2px solid #00A000]{
\frac{(mk)!}{k!^m}
{mk+1\choose m}.}$$
Concluding remark. The permutation group acting on the
children may sensibly be taken to be the one consisting of the
identity permutation, the cyclic group or the symmetric group. These
correspond to the operators $\textsc{SEQ},$ $\textsc{CYC}$ and
$\textsc{SET}.$ The cyclic case signifies that the tree is being
embedded in three-space and rooted at the origin so that we may rotate
the children. Children at a given depth are all in the same horizontal
plane parallel to the plane at $z=0$ and children of the same parent
are located on a circle in their plane separated by an angle of $2\pi
/k$ radians.
A recent comment by OP indicates what is being asked for is the
operator $\textsc{SEQ}$ (plane trees). We get in the labeled case
$$\bbox[5px,border:2px solid #00A000]{
[[n\equiv 1 \bmod k]]
(n-1)! {n\choose (n-1)/k}.}$$
These are not in the OEIS. I suspect this is because the labeled case
is obtained from the unlabeled case by trivially distributing the $n$
labels in $n!$ ways into the nodes of an unlabeled source tree. What
happens here is that the $\textsc{SEQ}$ operator is the same
algebraically in the labeled and the unlabeled case, which does not
hold for $\textsc{CYC}$ and $\textsc{SET}.$ The sequence operator
admits no symmetries, which is why all $n!$ distributions of the
labels are different, forming an orbit of order one.
