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There's an easy counterexample to your guess: Let $M^6 = \mathrm{SU}(3)/\mathbb{T}^2$, where $\mathbb{T}^2\subset\mathrm{SU}(3)$ is the maximal torus (for example, the diagonal subgroup). In that case, there is a 3-parameter family of non-isometric metrics on $M^6$ that are invariant under $\mathrm{SU}(3)$, so they are not unique up to a constant (or even non-constant) scalar factor.

I imagine that $M^6$ does not carry a metric whose isometry group has dimension greater than $8=\dim\mathrm{SU}(3)$, but I don't have a proof handy.

On the other hand, it is true that any metric on $\mathbb{CP}^n$ whose isometry group has dimension $n(n{+}2)$ must be isometric to a constant scalar multiple of the Fubini-Study metric. Here is one argument:

Suppose that a connected, compact group $G$ acts effectively on $\mathbb{CP}^n$. Then, by averaging, there exists a $G$-invariant metric $g$. Moreover, since $H^2_{dR}(\mathbb{CP}^n,\mathbb{R})\simeq\mathbb{R}$, it follows from the Hodge Theorem that there is a $g$-harmonic $2$-form $\omega$ that represents a generator of $H^2_{dR}(\mathbb{CP}^n,\mathbb{R})$, and it is unique up to constant multiples. Since $G$ is connected, it follows that it must leave $\omega$ fixed. Moreover, because of the structure of the cohomology ring of $\mathbb{CP}^n$, the top-degree form $\omega^n$ must represent a generator of $H^{2n}_{dR}(\mathbb{CP}^n,\mathbb{R})$. In particular, $\omega^n$ does not vanish identically.

Thus, there is a point $p\in\mathbb{CP}^n$ such that $\omega_p\in \Lambda^2(T^*_pM)$ is a 2-form of full rank. Now consider the stabilizer $G_p\subset G$ of $p$. Since $G$ acts by isometries and $\mathbb{CP}^n$ is connected, it follows that $G_p$ injects into $\mathrm{O}(T_pM)$ and, moreover, must leave $\omega_p$ fixed. In particular, $G_p$ must lie inside a subgroup of $\mathrm{O}(T_pM)$ that fixes a complex structure $J:T_pM\to T_pM$ and hence must have dimension at most $\dim \mathrm{U}(n) = n^2$. Thus, we have $$ \dim G = \dim G_p + \dim G{\cdot}p \le n^2 + 2n = n(n{+}2). $$ If equality holds, then $\dim G_p = n^2$ and $G{\cdot}p = \mathbb{CP}^n$. Since, $G$ acts transitively on $\mathbb{CP}^n$, it follows that $\omega$ is everywhere of full rank and, after scaling $\omega$ so that it has comass 1, we have that $\omega(u,v) = g(Ju,v)$ for a unique almost-complex structure $J$ on $\mathbb{CP}^n$. Moreover, since $G_p$ contains $-I\in\mathrm{O}(T_pM)$, it follows that there is an element of $G$ that fixes $p$ and reverses all $g$-geodesics through $p$. Since $G$ acts transitively on $\mathbb{CP}^n$, it follows that $(\mathbb{CP}^n,g)$ is a Riemannian symmetric space. Using the classification, it follows that $G\simeq \mathrm{SU}(n{+}1)$ and that the metric $g$ is, up to isometry, a constant multiple of the standard Fubini-Study metric.

There's an easy counterexample to your guess: Let $M^6 = \mathrm{SU}(3)/\mathbb{T}^2$, where $\mathbb{T}^2\subset\mathrm{SU}(3)$ is the maximal torus (for example, the diagonal subgroup). In that case, there is a 3-parameter family of non-isometric metrics on $M^6$ that are invariant under $\mathrm{SU}(3)$, so they are not unique up to a constant (or even non-constant) scalar factor.

I imagine that $M^6$ does not carry a metric whose isometry group has dimension greater than $8=\dim\mathrm{SU}(3)$, but I don't have a proof handy.

On the other hand, it is true that any Riemannian metric on $\mathbb{CP}^n$ whose isometry group has dimension at least $n(n{+}2)$ must be isometric to a constant scalar multiple of the Fubini-Study metric. Here is one argument:

Suppose that a connected, compact group $G$ acts effectively and smoothly on $\mathbb{CP}^n$. Then, by averaging, there exists a $G$-invariant metric $g$. Moreover, since $H^2_{dR}(\mathbb{CP}^n,\mathbb{R})\simeq\mathbb{R}$, it follows from the Hodge Theorem that there is a $g$-harmonic $2$-form $\omega$ that represents a generator of $H^2_{dR}(\mathbb{CP}^n,\mathbb{R})$, and it is unique up to constant multiples. Since $G$ is connected, it follows that it must leave $\omega$ fixed. Moreover, because of the structure of the cohomology ring of $\mathbb{CP}^n$, the top-degree form $\omega^n$ must represent a generator of $H^{2n}_{dR}(\mathbb{CP}^n,\mathbb{R})$. In particular, $\omega^n$ does not vanish identically.

Thus, there is a point $p\in\mathbb{CP}^n$ such that $\omega_p\in \Lambda^2(T^*_pM)$ is a 2-form of full rank. Consider the stabilizer $G_p\subset G$ of $p$. Since $G$ acts by isometries and $\mathbb{CP}^n$ is connected, $G_p$ injects into $\mathrm{O}(T_pM)$ by identifying $g\in G_p$ with $g'(p):T_pM\to T_pM$. Moreover, $G_p$ leaves $\omega_p$ fixed. Thus, $G_p$ must lie inside a subgroup of $\mathrm{O}(T_pM)$ that fixes a complex structure $J:T_pM\to T_pM$ and hence must have dimension at most $\dim \mathrm{U}(n) = n^2$. Now, we have $$ \dim G = \dim G_p + \dim G/G_p = \dim G_p + \dim G{\cdot}p \le n^2 + 2n = n(n{+}2). $$ If equality holds, then $\dim G_p = n^2$ and $\dim G{\cdot}p = 2n = \dim \mathbb{CP}^n$. Thus, the orbit $G{\cdot}p$ is both open and closed in $\mathbb{CP}^n$, so $G$ acts transitively on $\mathbb{CP}^n$. It follows that $\omega$ is everywhere of full rank and, after scaling $\omega$ so that it has comass 1, we have that $\omega(u,v) = g(Ju,v)$ for a unique almost-complex structure $J$ on $\mathbb{CP}^n$ that is preservd by $G_p$, which has the same dimension as the connected group $\mathrm{U}(g_p,J_p)\simeq \mathrm{U}(n)$. Thus, $G_p = \mathrm{U}(g_p,J_p)$. Since $G_p$ contains $-I\in\mathrm{U}(g_p,J_p)$, it follows that there is an element of $G$ that fixes $p$ and reverses all $g$-geodesics through $p$. Since $G$ acts transitively on $\mathbb{CP}^n$, it follows that $(\mathbb{CP}^n,g)$ is a Riemannian symmetric space. Using the classification, it follows that $G\simeq \mathrm{SU}(n{+}1)/Z$ (where $Z\simeq\mathbb{Z}_{n+1}$ is the center of $\mathrm{SU}(n{+}1)$) and that the metric $g$ is, up to isometry, a constant multiple of the standard Fubini-Study metric.

On the other hand, it may well be true that any metric on $\mathbb{CP}^n$ whose isometry group has dimension $n(n{+}2)$ must be isometric to a constant scalar multiple of the Fubini-Study metric. This is clearly true for $n=1$, and I see how to prove it for $n=2$, but I don't immediately see how to generalize that proof to work for all $n$.

A sketch of a proof for $n=2$ goes like this: Suppose that $G$ is a connected compact group that acts effectively on a connected, simply-connected $4$-manifold $M$ and has dimension at least $8$.

If $G$ were to fix a point $p$, then the assumption of effectivity would imply that $G$ embeds into $\mathrm{O}(T_pM)\simeq \mathrm{O}(4)$, which has dimension $6$, contradicting the assumption that $G$ has dimension at least $8$. Similarly, $G$ cannot have an orbit consisting of a finite number of points.

If $G$ were to have a $1$-dimensional orbit, then, by passing to a finite index subgroup, we could assume that $G$ has a circle as an orbit (remember that $G$ is compact) and preserves the orientation of the circle. But then there would be a closed subgroup $H$ of codimension $1$ that would fix a point on this orbit, and $H$ could have dimension at most $6$, so $G$ could have dimension at most $7$. So, no $1$-dimensional orbits.

By similar arguments, one can rule out 2-dimensional or $3$-dimensional orbits (left to the reader).

Thus, $G$ must have a $4$-dimensional orbit, and that must be all of $M$, i.e., $M = G/H$ where the closed subgroup $H$ is the stablizer of a point $p\in M$. Since $G$ and $M$ are connected and $M$ is simply-connected, the long exact sequence in homotopy implies that $H$ is connected as well. Now, $H$ embeds in $\mathrm{SO}(4)$ and must have dimension at least $4$. The only such connected subgroups of $\mathrm{SO}(4)$ are either conjugate to $\mathrm{U}(2)$ or equal to $\mathrm{SO}(4)$.

If $H$ were isomorphic to $\mathrm{SO}(4)$, then the metric on $M$ would have to have constant sectional curvature, implying that $M$ is diffeomorphic to the $4$-sphere.

If $H$ were isomorphic to $\mathrm{U}(2)$, then $G$ would indeed have dimension $8$. Under what circumstances could this happen? Well, if $M$ is diffeomorphic to $\mathbb{CP}^2$, then it has $\pi_2(\mathbb{CP}^2)\simeq \mathbb{Z}$, and since, $G$ is a Lie group, it has $\pi_2(G)=0$, moreover, $\pi_1(\mathrm{U}(2))\simeq \mathbb{Z}$, so the long exact sequence in homotopy gives $$ 0\longrightarrow\mathbb{Z}\longrightarrow \mathbb{Z}\longrightarrow\pi_1(G)\longrightarrow0. $$ In particular, $\pi_1(G)$ must be finite. Thus, the simply-connected cover of $G$ must be compact (since $G$ is compact), and hence it must be an $8$-dimensional product of compact simple groups. By Cartan's classification of compact simple, simply-connected Lie groups, this simply-connected cover can only be $\mathrm{SU}(3)$.

From here, it is now easy to argue that we must have $M=\mathrm{SU}(3)/\mathrm{S}(\mathrm{U}(1)\times\mathrm{U}(2))$, and this admits only one $\mathrm{SU}(3)$-invariant metric up to a constant scalar factor.

Remark: In order to carry this out for all $n$, basically you want to show that, if a compact Lie group of dimension at least $n(n{+}2)$ acts effectively on a connected, simply-connected $M^{2n}$, then it has to act transitively (by ruling out any lower dimensional orbits). Once you have that (and I think that maybe you can get this mostly by counting dimensions), I think that the above argument will go through to prove that either $(G,M) = (\mathrm{SO}(2n{+}1,S^{2n})$ or $G$ is $\mathrm{SU}(n{+}1)$ modulo its (finite) center and $M$ is $\mathbb{CP}^{n}$.

On the other hand, it is true that any metric on $\mathbb{CP}^n$ whose isometry group has dimension $n(n{+}2)$ must be isometric to a constant scalar multiple of the Fubini-Study metric. Here is one argument:

Suppose that a connected, compact group $G$ acts effectively on $\mathbb{CP}^n$. Then, by averaging, there exists a $G$-invariant metric $g$. Moreover, since $H^2_{dR}(\mathbb{CP}^n,\mathbb{R})\simeq\mathbb{R}$, it follows from the Hodge Theorem that there is a $g$-harmonic $2$-form $\omega$ that represents a generator of $H^2_{dR}(\mathbb{CP}^n,\mathbb{R})$, and it is unique up to constant multiples. Since $G$ is connected, it follows that it must leave $\omega$ fixed. Moreover, because of the structure of the cohomology ring of $\mathbb{CP}^n$, the top-degree form $\omega^n$ must represent a generator of $H^{2n}_{dR}(\mathbb{CP}^n,\mathbb{R})$. In particular, $\omega^n$ does not vanish identically.

Thus, there is a point $p\in\mathbb{CP}^n$ such that $\omega_p\in \Lambda^2(T^*_pM)$ is a 2-form of full rank. Now consider the stabilizer $G_p\subset G$ of $p$. Since $G$ acts by isometries and $\mathbb{CP}^n$ is connected, it follows that $G_p$ injects into $\mathrm{O}(T_pM)$ and, moreover, must leave $\omega_p$ fixed. In particular, $G_p$ must lie inside a subgroup of $\mathrm{O}(T_pM)$ that fixes a complex structure $J:T_pM\to T_pM$ and hence must have dimension at most $\dim \mathrm{U}(n) = n^2$. Thus, we have $$ \dim G = \dim G_p + \dim G{\cdot}p \le n^2 + 2n = n(n{+}2). $$ If equality holds, then $\dim G_p = n^2$ and $G{\cdot}p = \mathbb{CP}^n$. Since, $G$ acts transitively on $\mathbb{CP}^n$, it follows that $\omega$ is everywhere of full rank and, after scaling $\omega$ so that it has comass 1, we have that $\omega(u,v) = g(Ju,v)$ for a unique almost-complex structure $J$ on $\mathbb{CP}^n$. Moreover, since $G_p$ contains $-I\in\mathrm{O}(T_pM)$, it follows that there is an element of $G$ that fixes $p$ and reverses all $g$-geodesics through $p$. Since $G$ acts transitively on $\mathbb{CP}^n$, it follows that $(\mathbb{CP}^n,g)$ is a Riemannian symmetric space. Using the classification, it follows that $G\simeq \mathrm{SU}(n{+}1)$ and that the metric $g$ is, up to isometry, a constant multiple of the standard Fubini-Study metric.

Remark: In order to carry this out for all $n$, basically you want to show that, if a compact Lie group of dimension at least $n(n{+}2)$ acts effectively on a connected, simply-connected $M^{2n}$, then it has to act transitively (by ruling out any lower dimensional orbits). Once you have that (and I think that maybe you can get this mostly by counting dimensions), I think that the above argument will go through to prove that either $(G,M) = (\mathrm{SO}(2n{+}1,S^{2n})$ or $G$ is $\mathrm{SU}(n{+}1)$ modulo its (finite) center and $M$ is $\mathbb{CP}^{n}$.