I think they are true.

First of all, let's break out the diagonal as you suggested by writing $(-)^{\otimes n}:V\to V$ as the composite $V \xrightarrow{\Delta} V^n \xrightarrow{\otimes_n} V$. Since the 2-category of symmetric monoidal categories and lax symmetric monoidal functors has finite products, $F$ commutes with the $\Delta$'s (which are strict monoidal), so it suffices to show that $\otimes_n$ is monoidal and that we have a symmetric monoidal transformation (it doesn't make sense for a transformation to be "lax") $F \circ \otimes_n \to \otimes_n \circ F^n$.

Now, it's fairly straightforward to show that if $V$ is symmetric monoidal, then $\otimes : V\times V\to V$ is strong monoidal. By taking products with the identity and composing, we find that $\otimes_n : V^n \to V$ is also strong monoidal, and therefore $F\circ \otimes_n$ and $\otimes_n\circ F^n$ are lax monoidal.

More generally, if $X$ is a symmetric pseudomonoid in any 2-category with products, then $\otimes :X\times X\to X$ is strong monoidal, and hence so is $\otimes_n:X^n \to X$. But a lax symmetric monoidal functor can be identified with a symmetric pseudomonoid in the 2-category $\mathrm{Oplax}(\mathbf{2},\mathrm{Cat})$ whose objects are the arrows of Cat (functors) regarded as functors from the interval category $\mathbf{2}$ to Cat, and whose morphisms are oplax transformations (which here are just 2-cells fitting in a square). Similarly, a strong symmetric monoidal morphism in that 2-category can be identified with a symmetric monoidal transformation $G\circ F \to F'\circ H$ where $G$ and $H$ are strong monoidal and $F$ and $F'$ are lax monoidal. Thus, applying the general result about symmetric pseudomonoids to our $F$, we get the desired transformation.

I think they are true.

First of all, let's break out the diagonal as you suggested by writing $(-)^{\otimes n}:V\to V$ as the composite $V \xrightarrow{\Delta} V^n \xrightarrow{\otimes_n} V$. Since the 2-category of symmetric monoidal categories and lax symmetric monoidal functors has finite products, $F$ commutes with the $\Delta$'s (which are strict monoidal), so it suffices to show that $\otimes_n$ is monoidal and that we have a symmetric monoidal transformation (it doesn't make sense for a transformation to be "lax") $F \circ \otimes_n \to \otimes_n \circ F^n$.

Now, it's fairly straightforward to show that if $V$ is symmetric monoidal, then $\otimes : V\times V\to V$ is strong monoidal. By taking products with the identity and composing, we find that $\otimes_n : V^n \to V$ is also strong monoidal, and therefore $F\circ \otimes_n$ and $\otimes_n\circ F^n$ are lax monoidal.

More generally, if $X$ is a symmetric pseudomonoid in any 2-category with products, then $\otimes :X\times X\to X$ is strong monoidal, and hence so is $\otimes_n:X^n \to X$. But a lax symmetric monoidal functor can be identified with a symmetric pseudomonoid in the 2-category $\mathrm{Oplax}(\mathbf{2},\mathrm{Cat})$ whose objects are the arrows of Cat (functors) regarded as functors from the interval category $\mathbf{2}$ to Cat, and whose morphisms are oplax transformations (which here are just 2-cells fitting in a square). Similarly, a strong symmetric monoidal morphism in that 2-category can be identified with a symmetric monoidal transformation $G\circ F \to F'\circ H$ where $G$ and $H$ are strong symmetric monoidal and $F$ and $F'$ are lax symmetric monoidal. Thus, applying the general result about symmetric pseudomonoids to our $F$, we get the desired transformation.

I think they are true.

First of all, let's break out the diagonal as you suggested by writing $(-)^{\otimes n}:V\to V$ as the composite $V \xrightarrow{\Delta} V^n \xrightarrow{\otimes_n} V$. Since the 2-category of symmetric monoidal categories and lax symmetric monoidal functors has finite products, $F$ commutes with the $\Delta$'s (which are strict monoidal), so it suffices to show that $\otimes_n$ is monoidal and that we have a symmetric monoidal transformation (it doesn't make sense for a transformation to be "lax") $F \circ \otimes_n \to \otimes_n \circ F^n$.

Now, it's fairly straightforward to show that if $V$ is symmetric monoidal, then $\otimes : V\times V\to V$ is strong monoidal. By taking products with the identity and composing, we find that $\otimes_n : V^n \to V$ is also strong monoidal, and therefore $F\circ \otimes_n$ and $\otimes_n\circ F^n$ are lax monoidal.

More generally, if $X$ is a symmetric pseudomonoid in any 2-category with products, then $\otimes :X\times X\to X$ is strong monoidal, and hence so is $\otimes_n:X^n \to X$. But a lax symmetric monoidal functor can be identified with a pseudomonoid in the 2-category $\mathrm{Oplax}(\mathbf{2},\mathrm{Cat})$ whose objects are the arrows of Cat (functors) regarded as functors from the interval category $\mathbf{2}$ to Cat, and whose morphisms are oplax transformations (which here are just 2-cells fitting in a square). Similarly, a strong monoidal morphism in that 2-category can be identified with a symmetric monoidal transformation $G\circ F \to F'\circ H$ where $G$ and $H$ are strong monoidal and $F$ and $F'$ are lax monoidal. Thus, applying the general result about symmetric pseudomonoids to our $F$, we get the desired transformation.

I think they are true.

First of all, let's break out the diagonal as you suggested by writing $(-)^{\otimes n}:V\to V$ as the composite $V \xrightarrow{\Delta} V^n \xrightarrow{\otimes_n} V$. Since the 2-category of symmetric monoidal categories and lax symmetric monoidal functors has finite products, $F$ commutes with the $\Delta$'s (which are strict monoidal), so it suffices to show that $\otimes_n$ is monoidal and that we have a symmetric monoidal transformation (it doesn't make sense for a transformation to be "lax") $F \circ \otimes_n \to \otimes_n \circ F^n$.

Now, it's fairly straightforward to show that if $V$ is symmetric monoidal, then $\otimes : V\times V\to V$ is strong monoidal. By taking products with the identity and composing, we find that $\otimes_n : V^n \to V$ is also strong monoidal, and therefore $F\circ \otimes_n$ and $\otimes_n\circ F^n$ are lax monoidal.

More generally, if $X$ is a symmetric pseudomonoid in any 2-category with products, then $\otimes :X\times X\to X$ is strong monoidal, and hence so is $\otimes_n:X^n \to X$. But a lax symmetric monoidal functor can be identified with a symmetric pseudomonoid in the 2-category $\mathrm{Oplax}(\mathbf{2},\mathrm{Cat})$ whose objects are the arrows of Cat (functors) regarded as functors from the interval category $\mathbf{2}$ to Cat, and whose morphisms are oplax transformations (which here are just 2-cells fitting in a square). Similarly, a strong symmetric monoidal morphism in that 2-category can be identified with a symmetric monoidal transformation $G\circ F \to F'\circ H$ where $G$ and $H$ are strong monoidal and $F$ and $F'$ are lax monoidal. Thus, applying the general result about symmetric pseudomonoids to our $F$, we get the desired transformation.