Maybe someone else can explain the math behind it, but consider this quick demonstration: I generate five vectors, each 100 numbers long. Each of these vectors is on a different scale, so I standardize them (i.e., create z-scored variables). That is, the mean is zero and the standard deviation is 1 for each of these five latent construct variables:
set.seed(1839)
## create five different z-score variables that represent latent constructs
data <- data.frame(
latent_construct_1 = scale(rnorm(100, 10, 4)),
latent_construct_2 = scale(rnorm(100, 3, 18)),
latent_construct_3 = scale(rnorm(100, -5, 7)),
latent_construct_4 = scale(rnorm(100, 0, 8)),
latent_construct_5 = scale(rnorm(100, 20, 20))
)
Let's check to make sure they are actually z-scores:
> sapply(data, mean)
latent_construct_1 latent_construct_2 latent_construct_3 latent_construct_4 latent_construct_5
-2.203951e-16 1.634435e-17 1.400464e-17 -1.449145e-17 7.852226e-17
>
> sapply(data, sd)
latent_construct_1 latent_construct_2 latent_construct_3 latent_construct_4 latent_construct_5
1 1 1 1 1
So, now let's say we average all five of these together:
## make a mean of all of these latent constructs
data$mean_latent_construct <- rowMeans(data)
Is this new variable a z-score? We can check to see if the mean is zero and standard deviation is one:
> ## is the mean zero?
> mean(data$mean_latent_construct)
[1] -2.436148e-17
>
> ## is the standard deviation one?
> sd(data$mean_latent_construct)
[1] 0.4599126
The variable is not a z-score, because the standard deviation is not one. However, we could now z-score this mean variable. Let's do that and compare the distributions:
## z-score the mean latent construct
data$mean_latent_construct_z <- scale(data$mean_latent_construct)
## compare distributions
library(tidyverse)
data <- data %>%
select(mean_latent_construct, mean_latent_construct_z) %>%
gather(variable, value)
ggplot(data, aes(x = value, fill = variable)) +
geom_density(alpha = .7) +
theme_light()

The z-scored aggregate variable of z-scores looks a lot different from the aggregate variable of z-scores.
In short: No, a mean of z-scored variables is not a z-score itself.